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Probability
Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii) different days?
Solution:
Possible outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ekta are –
(T, T), (T, W), (T, TH), (T, F), (T, S),
(W, T), (W, W), (W, TH), (W, F), (W, S),
(TH, T), (TH, W), (TH, TH), (TH, F), (TH, S),
(F, T), (F, W), (F, TH), (F, F), (F, S),
(S, T), (S, W), (S, TH), (S, F), (S, S).
where T = Tuesday, W = Wednesday, Th = Thursday, F = Friday and S = Saturday.
∴ Total number of possible outcomes = 5 x 5 = 25
(i) Let A be the event of visiting a particular shop on the same day by two customers. Then, outcomes favorable to A are (T, T), (W, W), (TH, TH), (F, F) and (S, S)
∴ Favorable number of outcomes for same day = 5
Hence, required probability = 5/25 = 1/5.
(ii) Let A be the event of visiting a particular shop by two customers on consecutive days. Then, outcomes favorable to A are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).
∴ Favorable number of outcomes for consecutive days = 8
Hence, the required probability = 5/25 = 1/5.
(iii) Let A be the event of visiting a particular shop by two customers on different days. Then, outcomes favorable to A are excluding (T, T), (W, W), (TH, TH), (F, F) and (S, S). So, these are 25 – 5 = 20 in number;
∴ Favorable number of outcomes for different days = 20
Hence, required probability = 20/25 = 4/5
Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws :
What is the probability that the total score is (i) even? (ii) 6? (iii) at least 6?
Solution:
Complete table is as under :
Clearly total number of possible outcomes = 6 x 6 = 36
(i) Let A be the event of getting total score even. Then, outcomes favorable to A are 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8 and 12
∴ Favorable number of outcomes for even numbers = 18
Hence, P(A) = 18/36 = 1/2
(ii) Let A be the event of getting total score 6. Then, outcomes favorable to A are 6, 6, 6 and 6.
∴ Favorable number of outcomes for 6 = 4
Hence, P(A) = 4/36 = 1/9
(iii) Let A be the event of the total score is at least 6. Then, outcomes favorable to A are 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12.
∴ Favorable number of outcomes for at least 6 = 15
Hence, P(A) = 15/36 = 5/12
Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let there be x blue balls in the bag.
∴ Total number of balls in the bag = 5 + x
Now, p1 = Probability of drawing a blue ball
= x/(5+x)
p2 = Probability of drawing a red ball = 5/(5+x)
But it is given that p1 = 2p2.
So, x/(5+x) = 2 x {5/(5+x)}
or x = 10
Hence, there are 10 blue balls in the bag.
Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
There are 12 balls in the box. Out of these 12 balls, one can be chosen in 12 ways.
∴ Total number of possible outcomes = 12
There are x black balls out of which one can be chosen ir x ways.
∴ Favorable number of outcomes for black ball = x
Hence, p1 = P (getting a black ball) = x/12.
If 6 more black balls are put in the box, then
Total number of balls in the box = 12 + 6 = 18
Number of black balls in the box = x + 6
∴ p2 = P (getting a black ball) = (x+6)/18
It is given that p2 = 2p1
So, (x+6)/18 = 2 x (x/12)
or (x+6)/18 = x/6
or x + 6 = 3x
or 2x = 6
or x = 3
Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Solution:
There are 24 marbles in the jar, some are green and others are blue.
Total number of possible outcomes = 24 Let there be x green marbles.
∴ Favorable number of outcomes for green marbles = x
∴ P(G) = x/24 But, P(G) = 2/3 [Given]
So, x/24 = 2/3
or x = 2/3 x 24 = 16
or Number of green marbles = 16
∴ Number of blue marbles = 24 – 16 = 8.
