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Course: Class 11 > Unit 14

Lesson 4: Solutions to select NCERT problems

Select problems from miscellaneous exercise

Solutions to some NCERT problems.

In this article we will look at solutions of a few selected problems from miscellaneous exercise on chapter 16 of NCERT.

Problem 1:

A box contains

10

red marbles,

20

blue marbles and

30

green marbles.

5

marbles are drawn from the box, what is the probability that

(i) all will be blue?
(ii) at least one will be green?

Solution:

First let us look at number of possible outcomes. We can pick any

5

marbles out of total

60

marbles. So number of possible choices is

^{60} C_{5}

(i) Number of cases in which all

5

marbles are blue is

^{20} C_{5}

, i.e. when all marbles are picked from the group of

20

blue marbles.

P(all blue) = \frac{^{20} C_{5}}{^{60} C_{5}}

(ii) See that

P(at least one green) = 1 - P(no green)

Number of cases in which no green marble is picked is

^{30} C_{5}

, i.e. when all marbles are picked from the group of

30

red and blue marbles.

\begin{aligned} P(no green) & = \frac{^{30} C_{5}}{^{60} C_{5}} \\ P(at least one green) & = 1 - P(no green) \\ = 1 - \frac{^{30} C_{5}}{^{60} C_{5}} \end{aligned}

Problem 2:

In a certain lottery

10000

tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy

(a) one ticket
(b) two tickets
(c)

10

tickets.

Solution:

Out of the

10000

tickets,

10

are winning tickets where as

9990

are losing tickets.

(a) Suppose we buy one ticket. We don't win if this ticket belongs to the

9990

losing tickets. In this case,

P(not getting a prize) = \frac{9990}{10000} = \frac{999}{1000}

(b) Suppose we buy two tickets. We have a total of

^{10000} C_{2}

possible combinations of two tickets. We don't win if both of these ticket belong to the

9990

losing tickets. In this case,

P(not getting a prize) = \frac{^{9990} C_{2}}{^{10000} C_{2}}

10

tickets. We have a total of

^{10000} C_{10}

possible combinations of

10

tickets. We don't win if all of these tickets belong to the

9990

losing tickets. In this case,

P(not getting a prize) = \frac{^{9990} C_{10}}{^{10000} C_{10}}

Problem 3:

Out of

100

students, two sections of

40

and

60

are formed. If you and your friend are among the

100

students what is the probability that

(a) you both enter the same section?
(b) you both enter the different sections?

Solution:

First let us look at the total number of possible outcomes. In how many ways can we distribute

100

students into two groups of

40

and

60

For the first section, we have

^{100} C_{40}

different combinations. The remaining

60

students form part of the second section. So, total number of possible distributions is

^{100} C_{40} \times^{60} C_{60} =^{100} C_{40}

(a) Now in how many distributions are you and your friend in the same section?

Let us call the sections

A

(with

40

students) and

B

(with

60

students).

If you and your friend are in section

A

, the remaining

38

students can be chosen from the remaining

98

students in

^{98} C_{38}

ways. The rest

60

students form part of section

B

. Number of distributions is

^{98} C_{38} \times^{60} C_{60} =^{98} C_{38}

If you and your friend are in section

B

, the

40

students of section

A

can be chosen in

^{98} C_{40}

ways. The rest

58

students form part of section

B

. Number of distributions is

^{98} C_{40} \times^{58} C_{58} =^{98} C_{40}

Therefore total number of distributions in which you and your friend are in the same section is equal to

\begin{aligned} ^{98} C_{38} +^{98} C_{40} \\ = \frac{98!}{38! 60!} + \frac{98!}{40! 58!} \\ = \frac{98! \times 39 \times 40}{40! 60!} + \frac{98! \times 59 \times 60}{40! 60!} \\ = \frac{98!}{40! 60!} (39 \times 40 + 59 \times 60) \\ = \frac{98!}{40! 60!} \times 20 \times (39 \times 2 + 59 \times 3) \\ = \frac{98!}{40! 60!} \times 20 \times 255 \end{aligned}

We have simplified the expression above to help us with the calculation of probability which we will do next.

\begin{aligned} P(same section) & = \frac{^{98} C_{38} +^{98} C_{40}}{^{100} C_{40}} \\ = \frac{\frac{98!}{40! 60!} \times 20 \times 255}{\frac{100!}{40! 60!}} \\ = \frac{98! \times 20 \times 255}{100!} \\ = \frac{20 \times 255}{100 \times 99} \\ = \frac{17}{33} \end{aligned}

(b) Now see that

\begin{aligned} P(different sections) & = 1 - P(same section) \\ = 1 - \frac{17}{33} \\ = \frac{16}{33} \end{aligned}

Try to find $P(different sections)$ ‍ via the long method and see that it matches with what we got here!

Problem 4:

Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Solution:

Suppose three envelopes are placed on the table. Now three letters

X

Y

and

Z

are put just below these envelopes.

X Y Z

Suppose this is the correct order. That is

X

should go in the first envelope,

Y

should go in the second envelope and

Z

should go in the third envelope.

Now these letters can be arranged in

3! = 6

ways.

\begin{aligned} X Y Z \leftarrow \\ X Z Y \leftarrow \\ Y X Z \leftarrow \\ Y Z X \\ Z X Y \\ Z Y X \leftarrow \end{aligned}

See that in

4

cases out of

6

, at least one letter is in its proper envelope. So, required probability

= \frac{4}{6} = \frac{2}{3}

Problem 5:

A

and

B

are two events such that

P(A) = 0.54

P(B) = 0.69

and

P(A \cap B) = 0.35

Find (i)

P(A \cup B)

, (ii)

P (A^{'} \cap B^{'})

, (iii)

P (A \cap B^{'})

and (iv)

P (B \cap A^{'})

Solution:

(i)

\begin{aligned} P (A \cup B) & = P (A) + P (B) - P (A \cap B) \\ = 0.54 + 0.69 - 0.35 \\ = 0.88 \end{aligned}

(ii)

\begin{aligned} A^{'} \cap B^{'} & = (A \cup B)^{'} \\ ∴ P (A^{'} \cap B^{'}) & = P (A \cup B)^{'} \\ = 1 - P (A \cup B) \\ = 1 - 0.88 \\ = 0.12 \end{aligned}

(iii) First check out what

A \cap B^{'}

is.

Now see that

\begin{aligned} P (A \cap B^{'}) & = P (A) - P (A \cap B) \\ = 0.54 - 0.35 \\ = 0.19 \end{aligned}

(iv) First check out what

B \cap A^{'}

is.

Now see that

\begin{aligned} P (B \cap A^{'}) & = P (B) - P (A \cap B) \\ = 0.69 - 0.35 \\ = 0.34 \end{aligned}

Problem 6:

4

-digit numbers greater than

5000

are randomly formed from the digits

0

1

3

5

, and

7

, what is the probability of forming a number divisible by

5

when, (i) the digits are repeated? (ii) the repetition of digits is not allowed?

Solution:

(i) Repetition of digits allowed.

In this case, how many

4

digit numbers greater than

5000

can we form using

0

1

3

5

and

7

\underset{―}{5 / 7} \underset{―}{} \underset{―}{} \underset{―}{}

Because the numbers are greater than

5000

, we can only have

5

7

in the first place. So there are

2

choices for the first place.

Since repetition is allowed, we can fill the next places with any of the five digits:

0

1

3

5

and

7

. This gives us

5

choices for the second, third and fourth place.

Total possible numbers

= 2 \times 5 \times 5 \times 5 = 250

But this also includes the number

5000

which we have to discard. So, total possible numbers

= 249

Now how many of these are divisible by

5

\underset{―}{5 / 7} \underset{―}{} \underset{―}{} \underset{―}{0 / 5}

If the number is divisible by

5

, we can only have

0

5

in the last place. So we only have

2

choices for the last place.

Total numbers divisible by

5

is equal to

2 \times 5 \times 5 \times 2 = 100

. But again we discard the number

5000

which leaves us with

99

numbers.

Therefore, probability that a randomly formed number is divisible by

5

\frac{99}{249}

Note: NCERT gives the answer as $\frac{2}{5}$ ‍ or $\frac{100}{250}$ ‍. That would be the case if we include the number $5000$ ‍ as an allowed choice.

(ii) Repetition of digits not allowed.

Try it out

In this case, how many

4

digit numbers greater than

5000

can we form using

0

1

3

5

and

7

\underset{―}{5 / 7} \underset{―}{} \underset{―}{} \underset{―}{}

Again we have

2

choices for the first place. After filling the first place, we have used one of the five digits. That leaves us with

4

choices for the second place. After filling the second place, we are left with

3

choices for the third place, and similarly we have

2

choices for the fourth place.

Total possible numbers

= 2 \times 4 \times 3 \times 2 = 48

Try it out

How many of the above numbers are divisible by

5

\underset{―}{5} \underset{―}{} \underset{―}{} \underset{―}{0}

Suppose the number starts with

5

. It should end with

0

to be divisible by

5

. We cannot have $5$ ‍ in the last place because repetition is not allowed.

That leaves us with

3

choices for the second place and

2

choices for the third place. So number of choices for this case

= 1 \times 3 \times 2 \times 1 = 6

\underset{―}{7} \underset{―}{} \underset{―}{} \underset{―}{0 / 5}

Now, suppose the number starts with

7

. Now it can end in

0

5

to be divisible by

5

Two digits get used up for the first and last place. That leaves us with

3

choices for the second place and

2

choices for the third place. So number of choices for this case

= 1 \times 3 \times 2 \times 2 = 12

Total possible numbers divisible by

5

are

6 + 12 = 18

Therefore, probability that a randomly formed number is divisible by

5

\frac{18}{48} = \frac{3}{8}

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