# Probability : Exercise 15.2 (Optional) (Mathematics NCERT Class 10th) Q.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?
Sol.         Elementary events 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.
Therefore, Total number of elementary events = 5 $\times$ 5 = 25
(i) Let A be the event of visiting a particular shop on the same day by two customers.
Then, elementary events favourable to A are (T, T), (W, W), (TH, TH), (F, F) and (S, S)

Therefore, Favourable number of elementary events = 5
Hence, required probability = ${5 \over {25}} = {1 \over 5}$
(ii) Let A be the event of visiting a particular shop by two customer on consecutive days. Then, elementary even favourable to A are (T, W), (W, T), (W, TH), (TH, W) (TH, F) (F, TH) (S, F) (F, S).
Therefore, Favourable number of elementary events = 8
Hence, the required probability = ${8 \over {25}}$
(iii) Let A be the event of visiting a particular shop by two customers on different days. Then, elementary events favourable to A are excluding (T, T), (W, W), (TH, TH), (F, F) and (S, S). So, these are 25 – 5 = 20 in number
Therefore, Favourable number of elementary events = 20
Hence, required probability = ${{20} \over {25}} = {4 \over 5}$.

Q.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 time 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?
Sol.       Complete table is as under :
Number in first throw

Clearly total number of elementary events
= 6$\times$ 6 =36
(i) Let A be the event of getting total score even. Then, elementary events favourable to A are 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8 and 12
Therefore, Favourable number of elementary events = 18
Hence, P(A) = ${8 \over {36}} = {1 \over 2}$
(ii) Let A be the event of getting total score 6. Then, elementary events favourable to A are 6, 6, 6 and 6.
Therefore, Favourable number of elementary events = 4
Hence, P(A) = ${4 \over {36}} = {1 \over 9}$
(iii) Let A be the event of the total score is at least 6. Then, elementary events favourable to A are 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12  Therefore, Favourable number of elementary events = 15
Hence, P(A) = ${{15} \over {36}} = {5 \over {12}}$.

Q.3      A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball double that of a red ball, determine the number of blue balls in the bag.
Sol.         Let there be x blue balls in the bag.
Therefore, Total number of balls in the bag = 5 + x
Now, ${p_1}$ = Probability of drawing a blue ball
= ${x \over {5 + x}}$
${p_2}$ = Probability of drawing a red ball
=  ${5 \over {5 + x}}$
But it is given that ${p_1}$ = $2{p_2}$
$\Rightarrow {x \over {5 + x}} = 2 \times {5 \over {5 + x}}$
$\Rightarrow x = 10$
Hence, there are 10 blue balls in the bag.

Q.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.
Sol.       There are 12 balls in the box. Out of these 12 balls, one can be chosen in 12 ways.
Therefore, Total number of elementary events = 12
There are x black balls out of which one can be chosen in x ways.
Therefore, Favourable number of elementary events = x
Hence, ${p_1}$ = P (getting a black ball) = ${x \over {12}}$
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
Therefore, ${p_2}$ = P (getting a white ball) = ${{x + 6} \over {18}}$
It is given that ${p_2} = 2{p_1}$
$\Rightarrow {{x + 6} \over {18}} = 2{\rm{ \times }}{x \over {12}}$
$\Rightarrow {{x + 6} \over {18}} = {x \over 6}$
$\Rightarrow$ x + 6 = 3x
$\Rightarrow$ 2x = 6
$\Rightarrow$ x = 3

Q.5      A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the jar, the probability that it is green is ${2 \over 3}$. and the number of blue balls in the jar.
Sol.         There are 24 marbles in the jar, some are green and others are blue.
Therefore, Total number of elementary events = 24
Let there be x green falls
Therefore, Favourable number of elementary evens = x
Therefore, P(G) = ${x \over {24}}$ But, P(G) = ${2 \over 3}$                           [Given]
$\Rightarrow$ ${x \over {24}} = {2 \over 3}$
$\Rightarrow$ x = ${2 \over 3} \times 24$ = 16
$\Rightarrow$ Number of green marbles = 16
$\Rightarrow$ Number of blue marbles = 24 – 16 = 8

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