Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b)

I.

Question 1.

In the experiment of tossing a coin n times, if the variable X denotes the number of heads and P(X = 4), P(X = 5), P(X = 6) are in arithmetic progression then find n.

Solution:

X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\) (∵ a coin is tossed)

Hint: a, b, c are in A.P.

⇒ 2b = a + c (or) b – a = c – a

Given, P(X = 4), P(X = 5), P(X = 6) are in A.P.

⇒ 2 × 30(n – 4) = 5[30 + n^{2} – 9n + 20]

⇒ 12n – 48 = n^{2} – 9n – 50

⇒ n^{2} – 21n + 98 = 0

⇒ n^{2} – 14n – 7n + 98 = 0

⇒ n(n – 14) – 7(n – 14) = 0

⇒ (n – 7) (n – 14) = 0

⇒ n = 7 or 14

Question 2.

Find the minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8.

Solution:

Let n be the number of times a fair coin tossed

X denotes the number of heads getting

X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)

Given P(X ≥ 1) ≥ 0.8

⇒ 1 – P(X = 0) ≥ 0.8

⇒ P(X = 0) ≤ 0.2

⇒ \({ }^n C_o\left(\frac{1}{2}\right)^n \leq 0.2\)

⇒ \(\left(\frac{1}{2}\right)^n \leq \frac{1}{5}\)

The Maximum value of n is 3

Question 3.

The probability of a bomb hitting a bridge is \(\frac{1}{2}\) and three direct hits (not necessarily consecutive) are needed to destroy it. Find the minimum number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.

Solution:

Let n be the minimum number of bombs required and X be the number of bombs that hit the bridge, then

X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)

Now P(X ≥ 3) > 0.9

⇒ 1 – P(X < 3) > 0.9

⇒ P(X < 3) < 0.1

⇒ P(X = 0) + P(X = 1) + P (X = 2) < 0.1

By trial and error, we get n ≥ 9

∴ The least value of n is 9

∴ n = 9

Question 4.

If the difference between the mean and the variance of a binomial variate is \(\frac{5}{9}\) then, find the probability for the event of 2 successes, when the experiment is conducted 5 times.

Solution:

Given n = 5

Let p be the parameters of the Binomial distribution

Mean – Variance = \(\frac{5}{9}\)

∴ Probability of the event of 2 success = \(\frac{80}{243}\)

Question 5.

One in 9 ships is likely to be wrecked when they are set on the sail, when 6 ships are on the sail, find the probability for

(i) Atleast one will arrive safely

(ii) Exactly, 3 will arrive safely

Solution:

p = probability of ship to be wrecked = \(\frac{1}{9}\)

q = 1 – p

= 1 – \(\frac{1}{9}\)

= \(\frac{8}{9}\)

Number of ships = n = 6

Question 6.

If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4).

Solution:

Mean = np = 2.4 ………(1)

Variance = npq = 1.44 ……….(2)

Dividing (2) by (1)

\(\frac{n p q}{n p}=\frac{1.44}{2.4}\)

⇒ q = o.6 = \(\frac{3}{5}\)

p = 1 – q

= 1 – 0.6

= 0.4

= \(\frac{2}{5}\)

Substituting in (1)

n(0.4) = 2.4

⇒ n = 6

P(1 < X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)

= \({ }^6 C_2 q^4 \cdot p^2+{ }^6 C_3 q^3 \cdot p^3+{ }^6 C_4 q^2 \cdot p^4\)

Question 7.

It is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.

Solution:

p = probability of defective bulb = \(\frac{1}{10}\)

q = 1 – p

= 1 – \(\frac{1}{10}\)

= \(\frac{9}{10}\)

n = Number of bulbs in the sample = 20

P(X > 2) = 1 – P(X ≤ 2)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

Question 8.

On average, rain falls on 12 days every 30 days, find the probability that, the rain will fall on just 3 days of a given week.

Solution:

Given p = \(\frac{12}{30}=\frac{2}{5}\)

q = 1 – p

= 1 – \(\frac{2}{5}\)

= \(\frac{3}{5}\)

n = 7, r = 3

P(X = 3) = ^{n}C_{r} . q^{n-r} . p^{r}

= \({ }^7 C_3\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)

= \(\text { 35. }\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)

= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.

For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.

Solution:

Let n, p be the parameters of a binomial distribution

Mean (np) = 6 ……..(1)

and variance (npq) = 2 ……..(2)

then \(\frac{n p q}{n p}=\frac{2}{6}\)

⇒ q = \(\frac{1}{3}\)

∴ p = 1 – q

= 1 – \(\frac{1}{3}\)

= \(\frac{2}{3}\)

From (1) np = 6

n(\(\frac{2}{3}\)) = 6

∴ n = 9

The first two terms of the distribution are

P(X = 0) = \({ }^9 C_0\left(\frac{1}{3}\right)^9=\frac{1}{3^9}\)

and P(X = 1) = \({ }^9 C_1\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)=\frac{2}{3^7}\)

Question 10.

In a city, 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.

Solution:

Average number of accidents per day

λ = \(\frac{10}{50}=\frac{1}{5}\) = 0.2

The probability that there win be 3 or more accidents in a day

P(X ≥ 3) = \(\sum_{\mathrm{K}=3}^{\infty} \mathrm{e}^{-\lambda} \cdot \frac{\lambda^{\mathrm{K}}}{\mathrm{K} !}, \lambda=0.2\)

II.

Question 1.

Five coins are tossed 320 times. Find the frequencies of the distribution of a number of heads and tabulate the result.

Solution:

5 coins are tossed 320 times

Probability of getting a head on a coin

p = \(\frac{1}{2}\), n = 5

Probability of having x heads

Question 2.

Find the probability of guessing at least 6 out of 10 answers in (i) True or false type examination (ii) multiple choice with 4 possible answers.

Solution:

(i) Since the answers are in True or false type

probability of success p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)

probability of guessing at least 6 out of 10

P(X = 6) = \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10-6} \cdot\left(\frac{1}{2}\right)^6\)

= \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10}\)

(ii) Since the answers are multiple-choice with 4 possible answers

Probability of success p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)

Probability of guessing at least 6 out of 10

P(X = 6) = \({ }^{10} C_6\left(\frac{3}{4}\right)^{10-6}\left(\frac{1}{4}\right)^6\)

= \({ }^{10} C_6 \cdot \frac{3^4}{4^{10}}\)

Question 3.

The number of persons joining a cinema ticket counter in a minute has Poisson distribution with parameter 6. Find the probability that

(i) no one joins the queue in a particular minute

(ii) two or more persons join the queue in a minute.

Solution:

Here λ = 6

(i) Probability that no one joins the queune in a particular minute

P(X = 0) = \(\frac{\mathrm{e}^{-\lambda} \lambda^0}{0 !}=\mathrm{e}^{-6}\)

(ii) Probability that two or more persons join the queue in a minute