Question:
If in a binomial distribution \(n=4\), \(P(X=0)=\frac{16}{81}\), then \(P(X=4)\) equals
If in a binomial distribution \(n=4\), \(P(X=0)=\frac{16}{81}\), then \(P(X=4)\) equals
Show Hint
Endpoints in binomial use \(p^n\) and \(q^n\).
Updated On: Mar 23, 2026
- \(\frac1{16}\)
- \(\frac1{81}\)
- \(\frac1{27}\)
- \(\frac18\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: \[ P(X=0)=q^4=\frac{16}{81}\Rightarrow q=\frac23 \]
Step 2: \[ p=1-q=\frac13 \]
Step 3: \[ P(X=4)=p^4=\frac1{81} \]
Was this answer helpful?
0
0
Top BITSAT Mathematics Questions
- The solution of $\sin^{-1}\,x-\sin^{-1}\,2x=\pm\frac{\pi}{3}$ is
- If one root of the quadratic equation $a x^{2}+b x+c=0$ is equal to $n^{\text {th }}$ power of the other root, then the value of: $a^{\frac{n}{n-1}} C^{\frac{1}{n-1}}+c^{\frac{n}{n-1}} a^{\frac{1}{n-1}}$ is equal to
- $(x -1) (x^2 - 5x + 7) < (x -,1),$ then $x$ belongs to
- An ellipse has $OB$ as semi-minor axis, $F$ and $F$ are its foci and the $\angle FBF$, is a right angle. Then, the eccentricity of the ellipse is
- $2^{1/4}. 2^{2/8}. 2^{3/16}. 2^{4/32}......\infty$ is equal to-
View More Questions
Top BITSAT binomial distribution Questions
- In a binomial distribution, the mean is $4$ and variance is $3$. Then its mode is :
- In a binomial distribution, the mean is 4 and variance is 3. Then, its mode is:
- If in a binomial distribution n=4 and P(X=0)=dfrac1681, then P(X=4) equals
- In a binomial distribution, the mean is 4 and variance is 3. Then its mode is:
- A coin is tossed 7 times. Each time a man calls head. Find the probability that he wins the toss on more occasions.
View More Questions
Top BITSAT Questions
- Two point charges $-q$ and $+ q$ are located at point's $(0, 0, - a)$ and, $(0, 0, a)$ respectively. The electric potential at a point $(0, 9, z)$, where $z > a$ is
- Which of the following must be known in order to determine the power output of an automobile?
- A large drop of oil (density $0.8 \,g / cm ^{3}$ and viscosity $\eta_{0}$ ) floats up through a column of another liquid (density $1.2\, g / cm ^{3}$ and viscosity $\eta_{L}$ ). Assuming that the two liquids do not mix, the velocity with which the oil drop rises will depend on :
- A body is projected vertically upwards at time $ t = 0$ and it is seen at a height $H$ at time $t_1$ and $t_2$ second during its flight. The maximum height attained is ($g$ is acceleration due to gravity)
- At what point of a projectile motion, acceleration and velocity are perpendicular to each other ?
View More Questions