Exercise: 15.2 (Page No: 311)

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:

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5×5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = ⅕

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) = 8/25

(iii) P (both visiting on the different days) = 1-P (both visiting on the same day)

So, P (both visiting on the different days) = 1-(⅕) = ⅘

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:

The table will be as follows:

+122336
1233447
2344558
2344558
3455669
3455669
67889912

So, the total number of outcomes = 6×6 = 36

(i) E (Even) = 18

P (Even) = 18/36 = ½

(ii) E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

(iii) E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

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:

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x+5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ P (drawing a blue ball) = [x/(x+5)] ——–(i)

Similarly,

P (drawing a red ball) = [5/(x+5)] ——–(i)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

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:

Total number of black balls = x

Total number of balls = 12

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (getting black balls) = x/12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x+6

Now, P (getting black balls) = (x+6)/18 ——————-(ii)

It’s given that, the probability of drawing a black ball now is double of what it was before

(ii) = 2 × (i)

(x+6)/18 = 2 × (x/12)

x + 6 = 3x

2x = 6

∴ x = 3

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 ⅔. Find the number of blue balls

in the jar.

Solution:

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24-x

P(getting green marble) = x/24

From the question, x/24 = ⅔

So, the total green marbles = 16

And, the total blue marbles = 24-x = 8


Frequently Asked Questions on Chapter 15- Probability

Complete the following statements Probability of an event E + Probability of the event ‘not E’ = ___________ ?

Probability of an event E + Probability of the event ‘not E’ = 1.

Which of the following experiments have equally likely outcomes Explain A driver attempts to start a car. The car starts or does not start?

This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

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 an orange flavoured candy?

We know that the bag only contains lemon-flavoured candies. So, The no. of orange flavoured candies = 0 ∴ The probability of taking out orange flavoured candies = 0/1 = 0

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Let the event wherein 2 students having the same birthday be E Given, P(E) = 0.992 We know, P(E) + P(not E) = 1 Or, P(not E) = 1 – 0.992 = 0.008 ∴ The probability that the 2 students have the same birthday is 0.008

Tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game? why?

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

What is the probability of ‘not E’, if P(E) = 0.05?

We know that,

P(E) + P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1 – P(E)

Or, P(not E) = 1 – 0.05

Hence, P(not E) = 0.95

A box contains 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?

The Total no. of marbles = 5 + 8 + 4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red marbles= 5

P (red marbles) = 5/17 = 0.29

(ii) Total number of white marbles= 8

P (white marbles) = 8/17 = 0.47

(iii) Total number of green marbles = 4

P (green marbles) = 4/17 = 0.23

Hence, P (not green) = 1 – P (green marbles)

= 1 – (4/17)

= 0.77

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Chapter-15-Probability

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