Exercise 11.5-Additional Problems - Chapter 11 - Probability Distributions - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
In a Binomial distribution if $\mathrm{n}=5$ and $\mathrm{P}(\mathrm{X}=3)=2 \mathrm{P}(\mathrm{X}=2)$ find $\mathrm{p}$.
Solution:
$
\begin{aligned}
& \mathrm{P}(\mathrm{X}=x)=n \mathrm{C}_x p^x q^{n-x} \\
& \mathrm{P}(\mathrm{X}=3)=5 \mathrm{C}_3 p^3 q^2 \text { and } \mathrm{P}(\mathrm{X}=2)=5 \mathrm{C}_2 p^2 q^3 \\
& 5 \mathrm{C}_3 p^3 q^2=2\left(5 \mathrm{C}_2 p^2 q^3\right) \\
& \therefore p=2 q \\
& p=2(1-p) \Rightarrow 3 p=2 ; p=\frac{2}{3}
\end{aligned}
$
Question 2.
If the sum of mean and variance of a Binomial Distribution is $4.8$ for 5 trials find the distribution.
Solution:
$\mathrm{np}+\mathrm{npq}=4.8 \Rightarrow \mathrm{np}(1+\mathrm{q})=4.8$
$5 \mathrm{p}[1+(1-\mathrm{p})]=4.8$
$\mathrm{p}^2-2 \mathrm{p}+0.96=0 \Rightarrow \mathrm{p}=1.2,0.8$
$\therefore \mathrm{p}=0.8 ; \mathrm{q}=0.2[\because \mathrm{p}$ cannot be greater than 1$]$
$\therefore$ The Binomial distribution is $\mathrm{P}[\mathrm{X}=\mathrm{x}]=5 \mathrm{C}_{\mathrm{x}}(0.8)^{\mathrm{x}}(0.2)^{5-\mathrm{x}}, \mathrm{x}=0$ to 5
Question 3.
If on an average 1 ship out of 10 do not arrive safely to ports. Find the mean and the standard deviation of the ships returning safely out of a total of 500 ships.
Solution:

Probability of a ship arriving safely
$
\begin{gathered}
p=\frac{9}{10} \\
\therefore q=1-p=1-\frac{9}{10}=\frac{1}{10} ; n=500 \\
\text { Mean }=n p=500 \times \frac{9}{10}=450 \\
\text { Standard deviation }=\sqrt{n p q} \\
\sqrt{500 \times \frac{9}{10} \times \frac{1}{10}}=\sqrt{45}=\sqrt{9 \times 5}=3 \sqrt{5} \\
\text { Mean }=450, \text { Standard deviation }=3 \sqrt{5}
\end{gathered}
$
Question 4.
The overall percentage of passes in a certain examination is 80. If six candidates appear in the examination what is the probability that at least five pass the examination.
Solution:
Pass percentage $=80 \%$
$\therefore$ Probability of a candidate passing in the examination
$
p=\frac{80}{100}=\frac{4}{5}
$
So, $q=1-\frac{4}{5}=\frac{1}{5}$
Number of candidates $=n=6$
$\therefore$ Probability mass function is $\mathrm{P}(\mathrm{X}=x)=6 \mathrm{C}_x p^x q^{6-x}=6_{\mathrm{C}_x}\left(\frac{4}{5}\right)^x\left(\frac{1}{5}\right)^{6-x}$
$
\begin{aligned}
P(X \geq 5) & =P(X=5)+P(X=6) \\
& =6_{C_5}\left(\frac{4}{5}\right)^5\left(\frac{1}{5}\right)^1+6_{C_6}\left(\frac{4}{5}\right)^6\left(\frac{1}{5}\right)^0 \\
& =6\left(\frac{4^5}{5^6}\right)+1\left(\frac{4^6}{5^6}\right)=\frac{4^5}{5^6}\{6+4\}=\frac{1024}{5^6} \times 10 \\
& =\frac{2048}{5^5}=\frac{2048}{3125}
\end{aligned}
$

Question 5.
In a hurdle race a player has to cross 10 hurdles. The probability that he will clear each hurdle is $5 / 6$. What is the probability that he will knock down less than 2 hurdles?
Solution:
Probability that the hurdle is cleared $=p=\frac{5}{6}$
$
\therefore q=1-\frac{5}{6}=\frac{1}{6} ; n=10
$
Probability mass function is $\mathrm{P}(\mathrm{X}=x)=10^{\mathrm{C}}\left(\frac{5}{6}\right)^x\left(\frac{1}{6}\right)^{10-x}$
Out of 10 , when less than 2 hurdles are knocked $\Rightarrow$ that greater than 8 hurdles are cleared.
$
\begin{aligned}
& =\mathrm{P}(\mathrm{X}>8)=\mathrm{P}(\mathrm{X}=9)+\mathrm{P}(\mathrm{X}=10) \\
& =10_{\mathrm{C}_9}\left(\frac{5}{6}\right)^9\left(\frac{1}{6}\right)^1+10_{\mathrm{C}_{10}}\left(\frac{5}{6}\right)^{10} \\
& =10\left(\frac{5^9}{6^{10}}\right)+1\left(\frac{5^{10}}{6^{10}}\right) \\
& =\frac{1}{6^{10}}\left\{5^9[10+5]\right\}=\frac{5^9}{6^{10}}[15]
\end{aligned}
$