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Course: Class 11 math (India) > Unit 15

Lesson 4: Solutions to select NCERT problems

Select problems from exercise 16.3

Solutions to some problems of NCERT exercise.

In this article we will look at solutions of a few selected problems from exercise 16.3 of NCERT.

Problem 1:

A fair coin is tossed four times, and a person wins Re

1

for each head and loses Rs

1.50

for each tail that turns up.

From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Solution:

On each toss the person may either win or lose. The final amount the person gets depends on the number of times he wins or loses.

There are a total of

5

possible cases. Below $W$ ‍ and $L$ ‍ denote a win and loss respectively.

Case

I

4

wins.

Outcome is

WWWW

. Here the person wins Rs

4

in total.

Case

II

3

wins and

1

loss.

Outcomes are

WWWL

WWLW

WLWW

and

LWWW

. Here the person wins Rs

1.50

in total.

Case

III

2

wins and

2

losses.

Outcomes are

WWLL

WLWL

etc. Here the person loses Re

1

in total.

Try it out

The bigger question is, how many outcomes are possible for this case?

We can list out all the outcomes but a shortcut would be to use permutation theory.

Basically, here we are arranging

4

objects,

2

identical

Ws

and

2

identical

Ls

in a line. Number of ways to do that is

\frac{4!}{2! 2!} = 6

Case

IV

1

win and

3

losses.

Try it out

Using permutation theory, find out the number of outcomes in this case.

There are

4

outcomes:

LLLW

LLWL

LWLL

and

WLLL

. Here the person loses Rs

3.50

in total.

Case

V

4

losses.

Outcome is

LLLL

. Here the person loses Rs

6

in total.

Now how do we find the probability for each case?

P = \frac{No. of favourable outcomes}{Total no. of outcomes}

Total number of outcomes is the sum of number of outcomes in each case

= 1 + 4 + 6 + 4 + 1 = 16

Required probabilities are

Case	Result	No. of favourable outcomes	Probability
$I$ ‍	Win Rs $4$ ‍	$1$ ‍	$\frac{1}{16}$ ‍
$II$ ‍	Win Rs $1.50$ ‍	$4$ ‍	$\frac{1}{4}$ ‍
$III$ ‍	Lose Re $1$ ‍	$6$ ‍	$\frac{3}{8}$ ‍
$IV$ ‍	Lose Rs $3.50$ ‍	$4$ ‍	$\frac{1}{4}$ ‍
$V$ ‍	Lose Rs $6$ ‍	$1$ ‍	$\frac{1}{16}$ ‍

Problem 2:

In a lottery, a person choses six different natural numbers at random from

1

20

, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game.

[Hint: order of numbers is not important.]

Solution:

What is the number of possible outcomes in this case?

The person can pick any combination of

6

numbers from

20

numbers. Number of possible choices is

^{20} C_{6}

Out of these, only

1

is the winning combination. The person wins if he picks this exact combination.

Probability of winning

= \frac{1}{^{20} C_{6}} = \frac{1}{\frac{20!}{6! 14!}} = \frac{1}{38760}

Problem 3:

Events

E

and

F

are such that

P(not E or not F) = 0.25

. State whether

E

and

F

are mutually exclusive.

Solution:

Mutually exclusive events can never happen simultaneously. In other words, if events

E

and

F

are mutually exclusive,

P(E and F) = P (E \cap F) = 0

Given

P(not E or not F) = 0.25

How can we convert this into an equation about probability of

E and F

See that

\begin{aligned} not E or not F & = E^{'} \cup F^{'} \\ E^{'} \cup F^{'} & = (E \cap F)^{'} using DeMorgan’s law \end{aligned}

Therefore

\begin{aligned} P (E^{'} \cup F^{'}) & = P (E \cap F)^{'} \\ = 1 - P (E \cap F) ∵ P (A^{'}) = 1 - P (A) \end{aligned}

Given

P (E^{'} \cup F^{'}) = 0.75

. So

\begin{aligned} 0.75 & = 1 - P (E \cap F) \\ P (E \cap F) & = 0.25 \end{aligned}

Because

P (E \cap F) \neq 0

E

and

F

are not mutually exclusive events.

Problem 4:

The probability that a student will pass the final examination in both English and Hindi is

0.5

and the probability of passing neither is

0.1

. If the probability of passing the English examination is

0.75

, what is the probability of passing the Hindi examination?

Solution:

Let us make a Venn diagram of probabilities using given data.

P(Hindi) = 0.5 + ?

Because sum of all probabilities should equal

1

, we have

\begin{aligned} 0.25 + 0.5 + ? + 0.1 & = 1 \\ ? & = 0.15 \end{aligned}

Therefore

P(Hindi) = 0.5 + 0.15 = 0.65

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