Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks.
If A and B are such that $\text{P}(\text{A}'\cup\text{B}')=\frac{2}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then $\text{P}(\text{A}')+\text{P}(\text{B}')=$ ________.

Answer

If A and B are such that $\text{P}(\text{A}'\cup\text{B}')=\frac{2}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then $\text{P}(\text{A}')+\text{P}(\text{B}')=\frac{10}{9}.$ Solution:
Here, $\text{P}(\text{A}'\cup\text{B}')=\frac{2}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
$\text{P}(\text{A}'\cup\text{B}')=1-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=1-\frac{2}{3}=\frac{1}{3}$
$\because\text{P}(\text{A}')+\text{P}(\text{B}')=1-\text{P}(\text{A})+1-\text{P}(\text{B})$
$=2-\big[\text{P}(\text{A})+\text{P}(\text{B})\big]$
$=2-\big[\text{P}(\text{A}\cup\text{B})+\text{P}(\text{A}\cap\text{B})\big]$
$=2-\Big(\frac{5}{9}+\frac{1}{3}\Big)=\frac{10}{9}$

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Question 21 Mark

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If A and B are two events such that $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P}(\text{A})=\text{p},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{5}{9},$ then p = __________.

Answer

If A and B are two events such that $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{p},\text{P}(\text{A})=\text{p},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9},$ then p $=\frac{1}{3}.$Solution:
Here, $\text{P}(\text{A})=\text{p},\text{P}(\text{B})=\frac{1}{3}$ and $\text{P}(\text{A}\cup\text{B})=\frac{5}{9}$
$\because\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\text{p}$
$\Rightarrow\text{P}(\text{A}\cap\text{B})=\frac{\text{p}}{3}$
And $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\frac{5}{9}=\text{p}+\frac{1}{3}-\frac{\text{p}}{3}$
$\Rightarrow\frac{5}{9}-\frac{1}{3}=\frac{2\text{p}}{3}$
$\Rightarrow\frac{5-3}{9}=\frac{2\text{p}}{3}$
$\Rightarrow\text{p}=\frac{2}{9}\times\frac{3}{2}=\frac{1}{3}$

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Question 31 Mark

Fill in the blanks.
If X follows binomial distribution with parameters n = 5, p and P (X = 2) = 9, P (X = 3), then p = _________.

Answer

If X follows binomial distribution with parameters n = 5, p and P (X = 2) = 9, P (X = 3), then $\text{p}=\frac{1}{10}.$
Solution:
$\because\text{P}(\text{X}=2)=9,\text{P}(\text{X}=3)$ (where, n = 5 and q = 1 - p)
$\Rightarrow{^5}\text{C}_2\text{p}^{2}(1-\text{p})^{3}=9\cdot{^5}\text{C}_3\text{P}{^3}(1-\text{P})^2$
$\Rightarrow\frac{5!}{2!3!}\text{p}^2(1-\text{p})^3=9.\frac{5!}{3!2!}\text{p}(1-\text{p})^2$
$\Rightarrow\frac{\text{p}^2(1-\text{p})^3}{\text{p}^3(1-\text{p})^2}=9$
$\Rightarrow\frac{(1-\text{p})}{\text{p}}=9$
$\Rightarrow9\text{p}+\text{p}=1$
$\Rightarrow\text{p}=\frac{1}{10}$

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Question 41 Mark

Fill in the blanks.
Let X be a random variable taking values $x_1, x_2, ......, x_n$ with probabilities $p_1, p_2, ..., p_n,$ respectively. Then var (X) = ________.

Answer

Let X be a random variable taking values $x_1, x_2, ...,x_n$ with probabilities $p_1, p_2, ..., p_n$, respectively.
Then $\text{Var}(\text{X})=\sum\text{P}_{\text{i}}\text{x}_{\text{i}}^2-\Big(\sum\text{P}_{\text{i}}\text{x}_{\text{i}}\Big)^2.$
Solution:
$\text{Var}(\text{X})=(\text{X}^2)-\big[\text{E}(\text{X})\big]^2$
$=\sum_\limits{\text{i}=1}^\text{n}\text{X}^{2}\text{P}(\text{X})-\bigg[\sum_\limits{\text{i}=1}^\text{n}\text{XP}(\text{X})\bigg]^2$
$=\sum\text{P}_{\text{i}}\text{x}_{\text{i}}^2-\Big(\sum\text{P}_{\text{i}}\text{x}_{\text{i}}\Big)^2$

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Question 51 Mark

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Let A and B be two events. If $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P}(\text{A}),$ then A is _________ of B.

Answer

Let A and B be two events. If $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\text{P}(\text{A}),$ then A is independent of B.Solution:
$\because\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ $\Rightarrow\text{P}(\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ $\Rightarrow\text{P}(\text{A})\cdot\text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})$ So, A is independent of B.

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Maths Fill In The Blanks[1 Marks ]

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