If r . v X B (n=5, p=1 3 ) then P(2<X<4)= - Vidyadip

MCQ

If $r . v \cdot X \sim B\left(n=5, p=\frac{1}{3}\right)$ then $P(2<X<4)=$

A
$\frac{80}{243}$
B
$\frac{40}{243}$
C
$\frac{40}{343}$
D
$\frac{80}{343}$

✓

Answer

$\begin{aligned} & \text {(b) : Since, } n=5, p=\frac{1}{3} \\
& \therefore q=1-p=1-\frac{1}{3}=\frac{2}{3} \\
& \Rightarrow P(2<X<4)=P(X=3) \\
& ={ }^5 C_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2=\frac{10 \times 4}{3^5}=\frac{40}{243}\end{aligned}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Binomial Distribution→Binomial Distribution — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

For any $2 \times 2$ matrix, if $\text{A(adj A)}=\begin{bmatrix} 10 \ \ 0 \\ 0 \ \ 10 \end{bmatrix},$ then $|A|$ is equal to:

If $\sin \left(\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1}(x)\right)=1$, then $x=$

$\sin \left[3 \sin ^{-1}\left(\frac{1}{5}\right)\right]=$

The angle between the lines represented by $\sqrt{3} x y-y^2=0$ is

Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?

The values of $x$ for which the angle between $\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$is obtuse and the angle between $\vec{\text{b}}$ and the $z-$axis is acute and less than $\frac{\pi}{6}$ are:

If $\cos ^{-1} \frac{3}{5}-\sin ^{-1} \frac{4}{5}=\cos ^{-1} x$, then $x=$

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+\lambda \hat{j}+\hat{k}, \vec{c}=\hat{i}-\hat{j}+4 \hat{k}$ and $\vec{a} \cdot(\vec{b} \times \vec{c})=10$, then $\lambda$ is equal to

Cosine of the angle between the lines
$\begin{array}{l} \bar{r}=5 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}) \text { and } \\ \bar{r}=7 \hat{i}+2 \hat{j}+2 \hat{k}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) \text { is } \end{array}$

For an invertible matrix $A$ if $A($ adj $A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$, then $|A|$ is