Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability5 Marks
Question
If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).
✓
Answer
Let n and p the parameter of binomial distribution.
Given,
$\text{Mean = np}=4\dots(1)$
$\text{Variance = npq}=2\dots(2)$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$
$\text{p}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distributon is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}\dots(3)$
$\text{P(X}\geq5)$
$=\text{P(X}=5)+\text{P(X}=6)+\text{P(X}=7)+\text{P(X}=8)$
$=\text{ }^{8}\text{C}_5\big(\frac{1}{2}\big)^{5}\big(\frac{1}{2}\big)^{3}+\text{ }^{8}\text{C}_6\big(\frac{1}{2}\big)^{6}\big(\frac{1}{2}\big)^{2}+\text{ }^{8}\text{C}_7\big(\frac{1}{2}\big)^{7}\big(\frac{1}{2}\big)+\text{ }^{8}\text{C}_8\big(\frac{1}{2}\big)^{8}$
[Using equation (3)]
$=\frac{8\times7\times6}{3\times2}\big(\frac{1}{2}\big)^8+\frac{8\times7}{2}\big(\frac{1}{2}\big)^8+8\big(\frac{1}{2}\big)^8+\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^8\big[56+28+8+1\big]$
$=\frac{93}{256}$
$\text{P(X}\geq5)=\frac{93}{256}$
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