Probability
Exercise 15.1
Q.No. 1. Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ___________
(ii) The probability of an event that cannot happen is __________ . Such an event is called _______ .
(iii) The probability of an event that is certain to happen is __________ . Such an event is called ____________ .
(iv) The sum of the probabilities of all the elementary events of an experiment is ________ .
(v) The probability of an event is greater than or equal to __________ and less than or equal to _____ .
Solution:
(i) 1.
(ii) 0, Impossible event.
(iii) 1, Sure or Certain event.
(iv) 1
(v) 0, 1.
Q.No. 2. Which of the following experiments have equally likely outcomes ? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trail is made to answer a true - false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) Not Equally likely. There is more options expect 'start or does not start'.
(ii) Not Equally likely. There is more options expect 'shoot or misses the shot'.
(iii) Equally likely. There is no more options expect 'right or wrong'.
(iv) Not Equally likely. Baby can be transgender too.
Q.No. 3. Why is tossing a coin considered to be fair way of deciding which team should get the ball at the beginning of football game?
Solution:
Tossing a coin is an equally likely event.
Q.No. 4. Which of the following cannot be the probability of an event ?
(i) 2/3 (ii) - 1.5 (iii) 15% (iv) 0.7
Solution:
The probability of an event is greater than or equal to 'zero' or less than or equal to 'one'.
So, the correct option is (ii) - 1.5
Q.No. 5. If P(E) = 0.05, what is the probability of 'not E' ?
Solution:
We know that
P(E) + P(not E) = 1
Now putting the value of P(E)
0.05 + P(not E) = 1
∴ P(not E) = 1 - 0.05 = 0.95 Ans.
Q.No. 6. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy ?
(ii) a lemon flavoured candy ?
Solution:
(i)
From question it is clear that
The numbers of orange flavoured candy = 0
Let the number of candies = x
P(orange flavoured) = No of orange flavoured candies/ No of total candies in the bag
∴ P(orange flavoured) = 0/x = 0 Ans.
(ii)
From question it is clear that
Let the numbers of candies = x
The number of lemon flavoured candies = x
P(lemon flavoured) = No. of lemon flavoured candies/ No. of total candies in the bag
∴ P(lemon flavoured) = x/x = 1 Ans.
Q.No. 7. It is given that in a group of 3 students, the probability of 2 students not having the same birth day is 0.992. What is the the probability that the 2 students have the same birthday ?
Solution:
Both events are complement to each other.
Let E the event of 2 students having same birthday.
∴'not E' the event of 2 students not having same birthday.
P(not E) = 0.992
P(E) = ?
We know that
P(E) + P(not E) = 1
P(E) = 1 - P(not E)
P(E) = 1 - 0.992 = 0.008 Ans.
Q.No. 8. A bag contain 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ?(ii) not red ?
Solution:
(i)
Let R be the event 'the ball taken out is red'.
Now the number of possible outcomes = 8
The number of outcomes favourable to the event R = 3
So, P(R) = The number of outcomes favourable to the event R/ The number of possible outcomes
∴ P(R) = 3/8 Ans.
(ii)
We know that
P(R) + P(not R) = 1
P(not R) = 1 - P(R)
P(not R) = 1 - 3/8
or, P(not R) = (8 - 3)/3
∴ P(not R) = 5/8 Ans.
Q.No. 9. A box contain 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green ?
Solution:
Let R be the event 'the ball taken out is red', W be the event 'the ball taken out is white' and G be the event 'the ball taken out is green'.
Now, the number of possible outcomes = 5 + 8 +4 = 17
(i)
Q.No. 8. A bag contain 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red ?(ii) not red ?
Solution:
(i)
Let R be the event 'the ball taken out is red'.
Now the number of possible outcomes = 8
The number of outcomes favourable to the event R = 3
So, P(R) = The number of outcomes favourable to the event R/ The number of possible outcomes
∴ P(R) = 3/8 Ans.
(ii)
We know that
P(R) + P(not R) = 1
P(not R) = 1 - P(R)
P(not R) = 1 - 3/8
or, P(not R) = (8 - 3)/3
∴ P(not R) = 5/8 Ans.
Q.No. 9. A box contain 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green ?
Solution:
Let R be the event 'the ball taken out is red', W be the event 'the ball taken out is white' and G be the event 'the ball taken out is green'.
Now, the number of possible outcomes = 5 + 8 +4 = 17
(i)
The number of outcomes favourable to the event R = 5
So, P(R) = The number of outcomes favourable to the event R/ The number of possible outcome
or, P(R) = 5/17 Ans.
(ii)
The number of outcomes favourable to the event W = 8
So, P(W) = The number of outcomes favourable to the event W/ The number of possible outcomes
or, P(W) = 8/17 Ans.
(iii)
The number of outcomes favourable to the event G = 4
So, P(G) = The number of outcomes favourable to the event G/ The number of possible outcomes
or, P(G) = 4/17
We know that
P(G) + P(not G) = 1
or, P(not G) = 1 - P(G)
or, P(not G) = 1 - 4/17
or, P(not G) = (17 - 4)/17
or, P(not G) = 13/17 Ans.
Q.No. 10. A piggy bank contains hundred 50p coins, fifty Rs.1 coins, twenty Rs.2 coins and 10 Rs.5 coins. It is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be 50p coin ? (ii) will not be a Rs.5 coin ?
Solution:
Let A be the event 'the coins fallen out is 50p' and B be the event 'the coins fallen out is Rs.5'
Now, the number of possible outcomes = 100 + 50 + 20 + 10 = 180.
(i)
The number of outcomes favourable to the event A = 100
So, P(A) = The number of outcomes favourable to the event A/The number of possible outcomes
or, P(A) = 100/180
or, P(A) = 5/9 Ans.
(ii)
The number of outcomes favourable to the event B = 10
So, P(B) = The number of outcomes favourable to the event B / The number of possible outcomes
or, P(B) = 10/180
or, P(B) = 1/18
We know that
P(B) + P(not B) = 1
or, P(not B) = 1 - P(B)
or, P(not B) = 1 - 1/18
or, P(not B) = (18 - 1)/18
or, P(not B) = 17/18 Ans.
Q.No.11. Gopi buys fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male?
Solution:
Let A the event 'the fish taken out is male'.
Now, the number of possible outcomes = 5 + 8 = 13
The number of outcomes favourable to the event A = 5
So, P(A) = The number of favourable outcomes to the event A/The number of possible outcomes
or, P(A) = 5/13 Ans.
Q.No. 12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8 and these equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9 ?
Solution:
Let A be the event 'the number pointing to 8'. B the event 'the number pointing to an odd number'. C be the event 'the number pointing to a number greater than 2. D be the event ' the number pointing less than 9.
Now, the number of possible outcomes = 8.
(i)
The number of outcomes favourable to A = 1
So, P(A) = The number of outcomes favourable to A/ The number of possible outcomes
or, P(A) = 1/8 Ans.
(ii)
The number of outcomes favourable to B = 4
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 4/8
or, P(B) = 1/2 Ans.
(iii)
The number of outcomes favourable to C = 6
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 6/8
or, P(C) = 3/4 Ans.
(iv)
The number of outcomes favourable to D = 8
So, P(D) = The number of possible outcomes favourable to E/ The total number of possible outcomes
or, P(D) = 8/8
or, P(D) = 1 Ans.
Q.No.13. A die is thrown once. Find the probability of getting
(i) a prime number (ii) a number lying between 2 and 6; (iii) an odd number.
Solution:
Let A be the event 'the number getting is prime', B be the event 'the number getting is lying between 2 and 6 and C be the event 'the number getting is an odd number'.
Now, the total number of possible outcomes = 6
The number of outcomes favourable to A = 3 {2,3,5}
The number of outcomes favourable to B = 3 {3,4,5}
The number of outcomes favourable to C = 3 {1,3,5}
(i)
So, P(A) = The number of outcomes favourable to A/ The number of total possible outcomes
or, P(A) = 3/6 = 1/2 Ans.
(ii)
So, P(B) = The number of outcomes favourable to B/ The number of total possible outcomes
or, P(B) = 3/6 = 1/2 Ans.
(iii)
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 3/6 = 1/2 Ans.
Q.No. 14. One card is drawn from a well-suffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Solution:
Let A be the event 'getting a king card of red colour', B be the event'getting a face card', C be the event ' getting a red face card', D be the event' getting the jack of hearts' E be the event ' getting spade', and F be the event 'getting the queen of diamond'.
The total number of possible outcomes = 52
The number outcomes favourable to A = 2
The number of outcomes favourable to B = 12
The number of outcomes favourable to C = 6
The number of outcomes favourable to D = 1
The number of outcomes favourable to E = 13
The number of outcomes favourable to F = 1
(i)
So, P(A) = The number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 2/52 = 1/26 Ans.
(ii)
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 12/52 = 3/13 Ans.
(iii)
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 6/52 = 3/26 Ans.
(iv)
So, P(D) = The number of outcomes favourable to D/ The total number of possible outcomes
or, P(A) = 1/52 = 1/52 Ans.
(v)
So, P(E) = The number of outcomes favourable to E/ The total number of possible outcomes
or, P(E) = 13/52 = 1/4 Ans.
(vi)
So, P(F) = The number of outcomes favourable to F/ The total number of possible outcomes
or, P(F) = 1/52 = 1/52 Ans.
Q.No.15. Five cards - the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) if the queen is drawn and put aside, what is the probability that the second card picked up is (a) ace ? (b) queen ?
Solution:
(i)
The total number of possible outcomes = 5
Let A be the event ' the drawn card is queen'.
The number of outcomes favourable to A = 1
So, P(A) = The number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 1/5 Ans.
(ii)
(a)
The total number of possible outcomes = 4
Let B be the event ' the drawn card is ace'.
The number of outcomes favourable to B = 1
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 1/4 Ans.
(b)
The total number of possible outcomes = 4
Let C be the event ' the drawn card is queen'.
The number of outcomes favourable to B = 0
So, P(C) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 0/4 = 0 Ans.
Q.No. 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
The total number of possible outcomes = 12 + 132 = 144
Let A be the event 'the pen taken out is a good one'
The number of outcomes favourable to A = 132
So,
P(A) = The number of outcomes favourable to A/ The total number of possible outcome
or, P(A) = 132/144 = 11/12 Ans.
Q.No. 17(i). A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii). Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:
(i)
The total number of possible outcomes = 20
Let A be the event 'the drawn bulb is defective' .
The total number of outcomes favourable to A = 4
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 4/20 = 1/5 Ans.
(ii)
Let B be the event 'the drawn bulb is not defective'.
The total number of outcomes favorable to B = 15 {One correct bulb is already drawn and out of 16 and not replaced}
The total number of possible outcomes = 19 {One bulb is already drawn out of 20 and not replaced}
So,
P(B) = The total number of outcomes favourable to B/ The total number of possible outcomes = 15/19 Ans.
Q.No.18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5.
Solution:
The total number of possible outcomes = 90.
Let A be the event 'the disc drawn is a two digit number'. {10, 11, .................. 90}
The total number of outcomes favourable to A = 81
Let B be the event 'the disc drawn is a perfect square'. {1, 4, 9, 16, 25, 36, 49, 64, 81}
The total number of outcomes favourable to B = 9
Let C be the event ' the disc drawn is a number divisible by 5'. {5, 10, 15, ......... 90}
The total number of outcomes favourable to C = 18
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 81/90 = 9/10 Ans.
P(B) = The total number of outcomes favourable to B/The total number of possible outcomes.
or, P(B) = 9/90 = 1/10 Ans.
P(C) = The total number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 18/90 = 1/5 Ans.
Q.No.19. A child has a die whose six faces show the letters as given below
A B C D E A
The die is thrown once. What is the probability of getting (i) A ? (ii) D ?
Solution:
The total number of possible outcomes = 6
Let A be the event 'getting the letter A'.
The total number of outcomes favourable to the event A = 2
Let D be the event 'getting the letter D'
The total number of outcomes favourable to the event D = 1
(i)
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 2/6 = 1/3 Ans.
(ii)
So,
P(B) = The total number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 1/6 Ans.
Q.No.20. Suppose you drop a die at random on the rectangular region shown in Fig. What is probability that it will land inside the circle with diameter 1m ?
Solution:
The total number of possible outcomes = Area of rectangular region = 6m²
Let A be the event 'the die land inside the circular region'.
The total number of outcomes favourable to A = Area of the circular region = 𝛑(1/2)² = 𝛑/4m²
So,
P(A) = The total number of possible outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 𝛑/24 Ans.
Q.No.21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it ?
(ii) She will not buy it
Solution:
(i)
(ii)
Let A be the event ' the drawn pen by shopkeeper is good'.
The total number of outcomes favourable to A = 124
The total number of possible outcomes = 144
She will buy = P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 124/144 = 31/36 Ans.
She will not buy = 1 - P(A)
or, 1 - 31/36
or, (36 - 31)/36
or, 5/36 Ans.
Q.No. 22. Refer to Example 13.(i) Complete the following table:
(ii) A student argues that ' there are 11 possible outcomes 2,3,4,5,6,7,8,9,10,11 and 12. Therefore, each of them has a probability
Do you agree with this argument ? Justify your answer.
Solution:
The total number of possible outcomes = 36
The total number of outcomes favourable to sum on 2 dise is 3' = {(1,2), (2,1)} = 2
Probability of the event = 2/36 = 1/18
The total number of outcomes favourable to sum on 2 dise is 4 = {(1,3),(2,2),(3,1)} = 3
Probability of the event = 3/36 = 1/12 Ans.
The total number of outcomes favourable to sum on 2 dise is 5 = {(1,4),(2,3),(3,2),(4,1)} = 4
Probability of the event = 4/36 = 1/9 Ans.
The total number of outcomes favourable to sum on 2 dise is 6 = {(1,5),(2,4),(3,3),(4,2),(5,1)} = 5
Probability of the event = 5/36 Ans.
The total number of outcomes favourable to sum on 2 dise is 7 = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6
Probability of the event = 6/36 = 1/6 Ans.
The total number of outcomes favourable to sum on 2 dise is 9 = {(3,6),(4,5),(5,4),(6,3)} = 4
Probability of the event = 4/36 = 1/9 Ans.
The total number of outcomes favourable to sum on 2 dise is 10 = {(4,6),(5,5),(6,4)} = 3
Probability of the event = 3/36 = 1/12 Ans.
The total number of outcomes favourable to sum on 2 dise is 11 = {(5,6),(6,5)} = 2
Probability of the event = 2/36 = 1/18 Ans.
(ii)
No. These 11 events are not equally likely.
Q.No. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let A be the event 'getting three heads or three tails'.
The total number of outcomes favourable to the event A = 2. {(H,H,H), (T,T,T)}
The total number of possible outcomes = 8. {(H,H,H),(H,H,T) ................. (T,T,T)}
So,
P(A) = The total number of outcomes favourable to the event A/ The total number of possible outcomes
or, P(A) = 2/8 = 1/4
or, P(A) + P(not A) = 1
or, P(not A) = 1 - 1/4 = 3/4 Ans.
Q.No.24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time (ii) 5 will come up at least once ?
Solution:
Let A be the event '5 will come up either time'.
The total number of outcomes favourable to A = 11 {(5,1),(5,2),............(5,5),(5,6),(1,5)....... (6,5)}
The total number of possible outcomes = 6x6 = 36
(ii)
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 11/36 Ans.
(i)
P(A) + P(not A) = 1
or, P(not A) = 1 - 11/36
or, P(not A) = (36-11)/36 = 25/36 Ans.
Q.No.25. Which of the following arguments are correct and which are not correct ?
Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three are possible outcomes two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Solution:
(i)
Total number of possible outcomes when two coins are thrown simultaneously = {(H,H),(H,T),(T,H),(T,T)} = 4
Probability to get two heads = 1/4
Probability to get tails = 1/4
Probability to get one of each = 2/4 = 1/2
Hence, this argument is not correct.
(ii)
Total number of possible outcomes when a die is thrown = {1, 2 , 3, 4, 5, 6} = 6
Probability to get an odd number = 3/6 = 1/2
Hence, this argument is correct.
This Exercise is over.
So, P(R) = The number of outcomes favourable to the event R/ The number of possible outcome
or, P(R) = 5/17 Ans.
(ii)
The number of outcomes favourable to the event W = 8
So, P(W) = The number of outcomes favourable to the event W/ The number of possible outcomes
or, P(W) = 8/17 Ans.
(iii)
The number of outcomes favourable to the event G = 4
So, P(G) = The number of outcomes favourable to the event G/ The number of possible outcomes
or, P(G) = 4/17
We know that
P(G) + P(not G) = 1
or, P(not G) = 1 - P(G)
or, P(not G) = 1 - 4/17
or, P(not G) = (17 - 4)/17
or, P(not G) = 13/17 Ans.
Q.No. 10. A piggy bank contains hundred 50p coins, fifty Rs.1 coins, twenty Rs.2 coins and 10 Rs.5 coins. It is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be 50p coin ? (ii) will not be a Rs.5 coin ?
Solution:
Let A be the event 'the coins fallen out is 50p' and B be the event 'the coins fallen out is Rs.5'
Now, the number of possible outcomes = 100 + 50 + 20 + 10 = 180.
(i)
The number of outcomes favourable to the event A = 100
So, P(A) = The number of outcomes favourable to the event A/The number of possible outcomes
or, P(A) = 100/180
or, P(A) = 5/9 Ans.
(ii)
The number of outcomes favourable to the event B = 10
So, P(B) = The number of outcomes favourable to the event B / The number of possible outcomes
or, P(B) = 10/180
or, P(B) = 1/18
We know that
P(B) + P(not B) = 1
or, P(not B) = 1 - P(B)
or, P(not B) = 1 - 1/18
or, P(not B) = (18 - 1)/18
or, P(not B) = 17/18 Ans.
Q.No.11. Gopi buys fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male?
Solution:
Let A the event 'the fish taken out is male'.
Now, the number of possible outcomes = 5 + 8 = 13
The number of outcomes favourable to the event A = 5
So, P(A) = The number of favourable outcomes to the event A/The number of possible outcomes
or, P(A) = 5/13 Ans.
Q.No. 12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8 and these equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9 ?
Solution:
Let A be the event 'the number pointing to 8'. B the event 'the number pointing to an odd number'. C be the event 'the number pointing to a number greater than 2. D be the event ' the number pointing less than 9.
Now, the number of possible outcomes = 8.
(i)
The number of outcomes favourable to A = 1
So, P(A) = The number of outcomes favourable to A/ The number of possible outcomes
or, P(A) = 1/8 Ans.
(ii)
The number of outcomes favourable to B = 4
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 4/8
or, P(B) = 1/2 Ans.
(iii)
The number of outcomes favourable to C = 6
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 6/8
or, P(C) = 3/4 Ans.
(iv)
The number of outcomes favourable to D = 8
So, P(D) = The number of possible outcomes favourable to E/ The total number of possible outcomes
or, P(D) = 8/8
or, P(D) = 1 Ans.
Q.No.13. A die is thrown once. Find the probability of getting
(i) a prime number (ii) a number lying between 2 and 6; (iii) an odd number.
Solution:
Let A be the event 'the number getting is prime', B be the event 'the number getting is lying between 2 and 6 and C be the event 'the number getting is an odd number'.
Now, the total number of possible outcomes = 6
The number of outcomes favourable to A = 3 {2,3,5}
The number of outcomes favourable to B = 3 {3,4,5}
The number of outcomes favourable to C = 3 {1,3,5}
(i)
So, P(A) = The number of outcomes favourable to A/ The number of total possible outcomes
or, P(A) = 3/6 = 1/2 Ans.
(ii)
So, P(B) = The number of outcomes favourable to B/ The number of total possible outcomes
or, P(B) = 3/6 = 1/2 Ans.
(iii)
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 3/6 = 1/2 Ans.
Q.No. 14. One card is drawn from a well-suffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts (v) a spade (vi) the queen of diamonds
Solution:
Let A be the event 'getting a king card of red colour', B be the event'getting a face card', C be the event ' getting a red face card', D be the event' getting the jack of hearts' E be the event ' getting spade', and F be the event 'getting the queen of diamond'.
The total number of possible outcomes = 52
The number outcomes favourable to A = 2
The number of outcomes favourable to B = 12
The number of outcomes favourable to C = 6
The number of outcomes favourable to D = 1
The number of outcomes favourable to E = 13
The number of outcomes favourable to F = 1
(i)
So, P(A) = The number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 2/52 = 1/26 Ans.
(ii)
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 12/52 = 3/13 Ans.
(iii)
So, P(C) = The number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 6/52 = 3/26 Ans.
(iv)
So, P(D) = The number of outcomes favourable to D/ The total number of possible outcomes
or, P(A) = 1/52 = 1/52 Ans.
(v)
So, P(E) = The number of outcomes favourable to E/ The total number of possible outcomes
or, P(E) = 13/52 = 1/4 Ans.
(vi)
So, P(F) = The number of outcomes favourable to F/ The total number of possible outcomes
or, P(F) = 1/52 = 1/52 Ans.
Q.No.15. Five cards - the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) if the queen is drawn and put aside, what is the probability that the second card picked up is (a) ace ? (b) queen ?
Solution:
(i)
The total number of possible outcomes = 5
Let A be the event ' the drawn card is queen'.
The number of outcomes favourable to A = 1
So, P(A) = The number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 1/5 Ans.
(ii)
(a)
The total number of possible outcomes = 4
Let B be the event ' the drawn card is ace'.
The number of outcomes favourable to B = 1
So, P(B) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 1/4 Ans.
(b)
The total number of possible outcomes = 4
Let C be the event ' the drawn card is queen'.
The number of outcomes favourable to B = 0
So, P(C) = The number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 0/4 = 0 Ans.
Q.No. 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
The total number of possible outcomes = 12 + 132 = 144
Let A be the event 'the pen taken out is a good one'
The number of outcomes favourable to A = 132
So,
P(A) = The number of outcomes favourable to A/ The total number of possible outcome
or, P(A) = 132/144 = 11/12 Ans.
Q.No. 17(i). A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii). Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:
(i)
The total number of possible outcomes = 20
Let A be the event 'the drawn bulb is defective' .
The total number of outcomes favourable to A = 4
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 4/20 = 1/5 Ans.
(ii)
Let B be the event 'the drawn bulb is not defective'.
The total number of outcomes favorable to B = 15 {One correct bulb is already drawn and out of 16 and not replaced}
The total number of possible outcomes = 19 {One bulb is already drawn out of 20 and not replaced}
So,
P(B) = The total number of outcomes favourable to B/ The total number of possible outcomes = 15/19 Ans.
Q.No.18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two digit number (ii) a perfect square number (iii) a number divisible by 5.
Solution:
The total number of possible outcomes = 90.
Let A be the event 'the disc drawn is a two digit number'. {10, 11, .................. 90}
The total number of outcomes favourable to A = 81
Let B be the event 'the disc drawn is a perfect square'. {1, 4, 9, 16, 25, 36, 49, 64, 81}
The total number of outcomes favourable to B = 9
Let C be the event ' the disc drawn is a number divisible by 5'. {5, 10, 15, ......... 90}
The total number of outcomes favourable to C = 18
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 81/90 = 9/10 Ans.
P(B) = The total number of outcomes favourable to B/The total number of possible outcomes.
or, P(B) = 9/90 = 1/10 Ans.
P(C) = The total number of outcomes favourable to C/ The total number of possible outcomes
or, P(C) = 18/90 = 1/5 Ans.
Q.No.19. A child has a die whose six faces show the letters as given below
A B C D E A
The die is thrown once. What is the probability of getting (i) A ? (ii) D ?
Solution:
The total number of possible outcomes = 6
Let A be the event 'getting the letter A'.
The total number of outcomes favourable to the event A = 2
Let D be the event 'getting the letter D'
The total number of outcomes favourable to the event D = 1
(i)
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 2/6 = 1/3 Ans.
(ii)
So,
P(B) = The total number of outcomes favourable to B/ The total number of possible outcomes
or, P(B) = 1/6 Ans.
Q.No.20. Suppose you drop a die at random on the rectangular region shown in Fig. What is probability that it will land inside the circle with diameter 1m ?
Solution:
The total number of possible outcomes = Area of rectangular region = 6m²
Let A be the event 'the die land inside the circular region'.
The total number of outcomes favourable to A = Area of the circular region = 𝛑(1/2)² = 𝛑/4m²
So,
P(A) = The total number of possible outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 𝛑/24 Ans.
Q.No.21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it ?
(ii) She will not buy it
Solution:
(i)
(ii)
Let A be the event ' the drawn pen by shopkeeper is good'.
The total number of outcomes favourable to A = 124
The total number of possible outcomes = 144
She will buy = P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 124/144 = 31/36 Ans.
She will not buy = 1 - P(A)
or, 1 - 31/36
or, (36 - 31)/36
or, 5/36 Ans.
Q.No. 22. Refer to Example 13.(i) Complete the following table:
Do you agree with this argument ? Justify your answer.
Solution:
The total number of possible outcomes = 36
The total number of outcomes favourable to sum on 2 dise is 3' = {(1,2), (2,1)} = 2
Probability of the event = 2/36 = 1/18
The total number of outcomes favourable to sum on 2 dise is 4 = {(1,3),(2,2),(3,1)} = 3
Probability of the event = 3/36 = 1/12 Ans.
The total number of outcomes favourable to sum on 2 dise is 5 = {(1,4),(2,3),(3,2),(4,1)} = 4
Probability of the event = 4/36 = 1/9 Ans.
The total number of outcomes favourable to sum on 2 dise is 6 = {(1,5),(2,4),(3,3),(4,2),(5,1)} = 5
Probability of the event = 5/36 Ans.
The total number of outcomes favourable to sum on 2 dise is 7 = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6
Probability of the event = 6/36 = 1/6 Ans.
The total number of outcomes favourable to sum on 2 dise is 9 = {(3,6),(4,5),(5,4),(6,3)} = 4
Probability of the event = 4/36 = 1/9 Ans.
The total number of outcomes favourable to sum on 2 dise is 10 = {(4,6),(5,5),(6,4)} = 3
Probability of the event = 3/36 = 1/12 Ans.
The total number of outcomes favourable to sum on 2 dise is 11 = {(5,6),(6,5)} = 2
Probability of the event = 2/36 = 1/18 Ans.
(ii)
No. These 11 events are not equally likely.
Q.No. 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let A be the event 'getting three heads or three tails'.
The total number of outcomes favourable to the event A = 2. {(H,H,H), (T,T,T)}
The total number of possible outcomes = 8. {(H,H,H),(H,H,T) ................. (T,T,T)}
So,
P(A) = The total number of outcomes favourable to the event A/ The total number of possible outcomes
or, P(A) = 2/8 = 1/4
or, P(A) + P(not A) = 1
or, P(not A) = 1 - 1/4 = 3/4 Ans.
Q.No.24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time (ii) 5 will come up at least once ?
Solution:
Let A be the event '5 will come up either time'.
The total number of outcomes favourable to A = 11 {(5,1),(5,2),............(5,5),(5,6),(1,5)....... (6,5)}
The total number of possible outcomes = 6x6 = 36
(ii)
So,
P(A) = The total number of outcomes favourable to A/ The total number of possible outcomes
or, P(A) = 11/36 Ans.
(i)
P(A) + P(not A) = 1
or, P(not A) = 1 - 11/36
or, P(not A) = (36-11)/36 = 25/36 Ans.
Q.No.25. Which of the following arguments are correct and which are not correct ?
Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three are possible outcomes two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3
(ii) If a die is thrown, there are two possible outcomes an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.
Solution:
(i)
Total number of possible outcomes when two coins are thrown simultaneously = {(H,H),(H,T),(T,H),(T,T)} = 4
Probability to get two heads = 1/4
Probability to get tails = 1/4
Probability to get one of each = 2/4 = 1/2
Hence, this argument is not correct.
(ii)
Total number of possible outcomes when a die is thrown = {1, 2 , 3, 4, 5, 6} = 6
Probability to get an odd number = 3/6 = 1/2
Hence, this argument is correct.
This Exercise is over.
