Solve the Following Question.(4 Marks)

Question 14 Marks

It is observed that it rains $12$ days out of $30$ days. Find the probability that
(i) it rains exactly $3$ days of the week.
(ii) it will rain at least $2$ days of a given week.

Answer

Let $X=$ the number of days it rains in a week.
$\mathrm{p}=$ probability that it rains
$\therefore p=\frac{12}{30}=\frac{2}{5}$
and $q=1-p=1-\frac{2}{5}=\frac{3}{5}$
Given : $n=7$
i.e. $X \sim B\left(7, \frac{2}{5}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $p(x)={ }^7 C_x\left(\frac{2}{5}\right)^x\left(\frac{3}{5}\right)^{7-x}, x=0,1,2, \ldots, 7$
$ \text { (i) } P(\text { it rains exactly } 3 \text { days of week })=P(X=3)$
$= p(3)={ }^7 C_3\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^{7-3}$
$= \frac{7 \times 6 \times 5}{3 \times 2 \times 1}\left(\frac{8}{125}\right)\left(\frac{81}{625}\right)$
$= 35\left(\frac{8}{125}\right)\left(\frac{81}{625}\right)=\frac{35 \times 8 \times 81}{5^7}$
$= \frac{22680}{78125}=0.2903 $
Hence, the probability that it rains exactly 3 days of week $=0.2903$.
$ \text { (ii) } P(\text { it will rain at least } 2 \text { days of the given week) }$
$=P(X \geqslant 2)=1-P(X<2)$
$=1-[P(X=0)+P(X=1)]$
$= 1-\left[{ }^7 C_0\left(\frac{2}{5}\right)^0\left(\frac{3}{5}\right)^{7-0}+{ }^7 C_1\left(\frac{2}{5}\right)^1\left(\frac{3}{5}\right)^{7-1}\right]$
$=1-\left[1(1)\left(\frac{3}{5}\right)^7+7\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)^6\right]$
$=1-\left[\frac{3}{5}+\frac{14}{5}\right]\left(\frac{3}{5}\right)^6$
$=1-\left(\frac{17}{5}\right)\left(\frac{729}{(5)^6}\right)=1-\frac{12393}{5^7}$
$=1-\frac{12393}{78125}=1-0.1586$
$=0.8414 $
Hence, the probability that it rains at least 2 days of a given week $=0.8414$

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Question 24 Marks

A computer installation has 10 terminals. Independently, the probability that anyone terminal will require attention during a week is 0.1. Find the probabilities that (i) 0 (ii) 1 (iii) 2 (iv) 3 or more, terminals will require attention during the next week.

Answer

Let $\mathrm{X}=$ number of terminals which required attention during a week.
$\mathrm{p}=$ probability that any terminal will require attention during a week
$\therefore p=0.1$ and $q=1-p=1-0.1=0.9$
Given: $\mathrm{n}=10$
$\therefore X \sim \mathrm{B}(10,0.1)$
The p.m.f. of $X$ is given by
$\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $\mathrm{p}(\mathrm{x})={ }^{10} C_x(0.1)^x(0.9)^{10-x}, \mathrm{x}=0,1,2, \ldots, 10$
(i) $\mathrm{P}$ (no terminal will require attention $)=\mathrm{P}(\mathrm{X}=0)$
$ =p(0)={ }^{10} \mathrm{C}_0(0.1)^0(0.9)^{10-0}$
$=1 \times 1 \times(0.9)^{10}=(0.9)^{10} $
Hence, the probability that no terminal requires attention $=(0.9)^{10}$
(ii) $\mathrm{P}(1$ terminal will require attention $)$
$ P(X=1)=p(1)={ }^{10} C_1(0.1)^1(0.9)^{10-1}$
$=10(0.1)(0.9)^9$
$=(1.0)(0.9)^9$
$ =(0.9)^9 $
Hence, the probability that 1 terminal requires attention $=(0.9)^9$.
(iii) $\mathrm{P}(2$ terminals will require attention)
$ P(X=2)=p(2)={ }^{10} \mathrm{C}_2(0.1)^2(0.9)^{10-2}$
$=\frac{10 \times 9}{1 \times 2}(0.1)^2(0.9)^8$
$=45(0.01)(0.9)^8$
$=(0.45) \times(0.9)^8 $
Hence, the probability that 2 terminals require attention $=(0.45)(0.9)^8$.
(iv) $\mathrm{P}$ (3 or more terminals will require attention)
$ =P(X \geqslant 3)$
$=1-P(x<3)$
$=1-[P(X=0)+P(X=1)+P(X=2)]$
$=1-[p(0)+p(1)+p(2)]$
$=1-\left[(0.9)^{10}+(0.9)^9+(0.45)(0.9)^8\right]$
$=1-\left[(0.9)^2+(0.9)^1+0.45\right](0.9)^8$
$=1-[0.81+0.9+0.45](0.9)^8$
$=1-(2.16) \times(0.9)^8 $
Hence, the probability that 3 or more terminals require attention $=1-(2.16) \times(0.9)^8$.

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Question 34 Marks

If a fair coin is tossed 10 times and the probability that it shows heads (i) 5 times (ii) in the first four tosses and tail in the last six tosses.

Answer

Let $X=$ number of heads.
$\mathrm{p}=$ probability that coin tossed shows a head
$ \therefore p=\frac{1}{2}$
$q=1-p=1-\frac{1}{2}=\frac{1}{2} $
Given: $n=10$
$\therefore \mathrm{X} \sim \mathrm{B}\left(10, \frac{1}{2}\right)$
The p.m.f. of $X$ is given by
$ \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^n C_x P^x q^{n-x}$
$\therefore p(x)={ }^{10} \mathrm{C}_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{10-x}={ }^{10} \mathrm{C}_x\left(\frac{1}{2}\right)^{10}$
$\quad x=0,1,2, \ldots 10 $
$\quad x=0,1,2, \ldots 10$
$ \text { (i) } \mathrm{P} \text { (coin shows heads } 5 \text { times) }=P[X=5]$
$=p(5)={ }^{10} \mathrm{C}_5 \cdot\left(\frac{1}{2}\right)^{10}$
$=\frac{10 !}{5 ! 5 !} \times \frac{1}{1024}$
$=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1} \times \frac{1}{1024}=\frac{63}{256}$
Hence, the probability that can shows heads exactly 5 times $=\frac{63}{256}$
(ii) $P$ (getting heads in first four tosses and tails in last six tosses $)=P(X=4)$
$=p(4) ={ }^{10} \mathrm{C}_4 \cdot\left(\frac{1}{2}\right)^{10}$
$ =\frac{10 \times 9 \times 8 \times 7 \times 6 !}{6 ! \times 4 \times 3 \times 2 \times 1} \times \frac{1}{1024}$
$ =210 \times \frac{1}{1024}=\frac{105}{512}$
Hence, the probability that getting heads in first four tosses and tails in last six tosses $=\frac{105}{512}$.

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Question 44 Marks

Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Answer

Let $X=$ number of sixes.
$p=$ probability that a die shows six in a single throw
$\therefore \mathrm{p}=\frac{1}{6}$
and $q=1-p=1-\frac{1}{6}=\frac{5}{6}$
Given: $n=6$
$\therefore \mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{6}\right)$
The p.m.f, of $X$ is given by
$ P(X=x)={ }^n \mathrm{C}_x p^x q^{n-x}$
$\text { i.e. } p(x)={ }^6 \mathrm{C}_x\left(\frac{1}{6}\right)^x\left(\frac{5}{6}\right)^{6-x}, x=0,1,2, \ldots, 6$
$P(\text { at most } 2 \text { sixes })=P[X \leqslant 2]$
$=p(0)+p(1)+p(2)$
$={ }^6 \mathrm{C}_0\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{6-0}+{ }^6 \mathrm{C}_1\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{6-1}+ $
$\quad{ }^6 \mathrm{C}_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{6-2}$
$ =1 \times 1 \times\left(\frac{5}{6}\right)^6+6 \times\left(\frac{1}{6}\right) \times\left(\frac{5}{6}\right)^5+\frac{6 !}{2 ! 4 !} \times\left(\frac{1}{6}\right)^2 \times\left(\frac{5}{6}\right)^4$
$=\left(\frac{5}{6}\right)^6+\left(\frac{5}{6}\right)^5+\frac{6 \times 5}{2 \times 1}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^4$
$=\left(\frac{5}{6}\right)^6+\left(\frac{5}{6}\right)^5+15 \times \frac{1}{36} \times\left(\frac{5}{6}\right)^4$
$=\left[\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)+\frac{15}{36}\right]\left(\frac{5}{6}\right)^4$
$=\left(\frac{25}{36}+\frac{5}{6}+\frac{15}{36}\right) \cdot\left(\frac{5}{6}\right)^4$
$=\left(\frac{25+30+15}{36}\right)\left(\frac{5}{6}\right)^4$
$=\frac{70}{36}\left(\frac{5}{6}\right)^4$
$=\frac{7}{3} \times \frac{10}{12} \times\left(\frac{5}{6}\right)^4$
$=\frac{7}{3} \times \frac{5}{6} \times\left(\frac{5}{6}\right)^4=\frac{7}{3}\left(\frac{5}{6}\right)^5 $
Hence, probability of throwing at most 2 sixes $=\frac{7}{3}\left(\frac{5}{6}\right)^5$.

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Maths Solve the Following Question.(4 Marks)

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