If in a binomial distribution n=4,P(X=0)=16 81 , then P(X=4) equa… - Vidyadip

Question

If in a binomial distribution $\text{n}=4,\text{P(X}=0)=\frac{16}{81},$ then $\text{P(X}=4)$ equals:

✓

Answer

$\frac{1}{81}$

Solution:
Given $\text{n}=4,\text{P(X}=0)=\frac{16}{81}$
$\text{P(X}=0)=\frac{16}{81}$
$\text{ }^5\text{C}_0\text{p}^0\text{q}^4=\frac{16}{81}$
$\text{q}^4=\frac{16}{81}$
$\text{q}=\frac{2}{3}\Rightarrow\text{p}=\frac{1}{3}$
$\Rightarrow\text{P(X}=4)=\text{ }^5\text{C}_4\big(\frac{1}{4}\big)^4=\frac{1}{81}$

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 "M.C.Q (1 Marks)" from Probability→Probability — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If a matrix $A$ is both symmetric and skew $-$ symmetric, then:

If $y=5 \cos x-3 \sin x$, then $\frac{d^2 y}{d x^2}$ is equal to

Evaluate: $\int \frac{x-4}{(x-2)^3} \cdot e^x d x$

The altitude of a cone is 20cm and its semi-vertical angle is 30°. If the semi-vertical angle is increasing at the rate of 2° per second, then the radius of the base is increasing at the rate of:

$30\text{cm}/\text{sec}.$
$\frac{160}{3}\text{cm}/\text{sec}.$
$10\text{cm}/\text{sec}.$
$160\text{cm}/\text{sec}.$

If $\text{A}=\begin{bmatrix}1&\text{a}\\0&1\end{bmatrix},$ then $A^n ($where $n \in N)$ equals:

The inverse of the function $\text{y}=\frac{10^\text{x}-10^{-\text{x}}}{10^\text{x}+10^{-\text{x}}}$ is:

$\log_{10}(2-\text{x})$
$\frac{1}{2}\log_{10}\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\frac{1}{2}\log_{10}(2\text{x}-1)$
$\frac{1}{4}\log\big(\frac{2\text{x}}{2-\text{x}}\Big)$

A straight line passes through (1, -2, 3) and perpendicular to the plane 2x + 3y - z = 7. Find the direction ratios of normal to plane:

If $\frac{d y}{d x}=\frac{x+y}{x}, y(1)=1$, then $y=$

If the binary operation $\odot$ is defined on the set $Q^+$ of all positive rational numbers by $\text{a}\odot\text{b}=\frac{\text{ab}}4$. Then, $3\odot\Big(\frac{1}5\odot\frac{1}2\Big)$ is equal to:

If $A=\left[\begin{array}{rrr}0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0\end{array}\right]$, then $A+2 A^T$ equals