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**NCERT TEXTBOOK QUESTIONS SOLVED**

**Page No. 311****EXERCISE 15.2**

**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.** Here, the number of all the possible outcomes

= 5 × 5 = 25

(i) For both customers visiting same day:

Number of favourable outcomes = 5

∴ Required probability = 1/25 = 1/5

(ii) For both the customers visiting on consecutive days:

Number of outcomes are: (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Wed., Tue.), (Thu., Wed.), (Fri., Thu.)

∴ Number of favourable outcomes = 8

⇒ Required probability = 8/25

(iii) For both the customers visiting on different days: We have probability for both visiting same day = 1/5

∴ Probability for both visiting on different days

=1 − [Probability for both visiting on the same day]

⇒ The required probability = 4/5.**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 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?**

∴ Number of all possible outcomes = 36

Favourable outcomes = 18

[a The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8]

∴ The required probability = 18/36 = 1/2.

In list of score, we have four 6′ s.

∴ Favourable outcomes = 4

∴ Required probability = 4/36 = 1/9

The favourable scores are: 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12

∴ Number of favourable outcomes = 15

⇒ Required probability = 15/36 = 5/12.

∴ Total number of balls = x + 5

Number of possible outcomes = (x + 5).

For a blue ball favourable outcomes = x

Similarly, probability of drawing a red ball

Now, we have

Thus the required number of blue balls = 10.

∴ Number of possible outcomes = 12

Case-I: For drawing a black ball

Number of favourable outcomes = x

∴ Probability of getting a black ball = x/12.

Case-II: When 6 more black balls are added

Now, the total number of balls

= 12 + 6

= 18

⇒ Number of possible outcomes = 18

Since, the number of black balls now =(x + 6).

⇒ Number of favourable outcomes = (x + 6)

∴ Required probability

Applying the given condition:

Thus, the required value of x is 3.

∴ Number of possible outcomes = 24.

Let there are x blue marbles in the jar.

∴ Number of green marbles = 24 − x

⇒ Favourable outcomes = (24 − x)

∴ Required probability for drawing a green marble

Now, according to the condition, we have:

Thus, the required number of blue balls is 8.

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