Let in a Binomial distribution,consisting of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure.Given $n = 5$ independent trials.The probability of $X$ succes

Vedclass · Accepted Answer

$\frac{32}{625}$

Vedclass · Answer

(C) Let $p$ be the probability of success and $q = 1 - p$ be the probability of failure.Given $n = 5$ independent trials.The probability of $X$ successes is given by $P(X=k) = {}^nC_k p^k q^{n-k}$.We have $P(X=1) = 0.4096$ and $P(X=2) = 0.2048$.${}^5C_1 p^1 q^4 = 5pq^4 = 0.4096$ (Equation $1$).${}^5C_2 p^2 q^3 = 10p^2q^3 = 0.2048$ (Equation $2$).Dividing Equation $2$ by Equation $1$:$\frac{10p^2q^3}{5pq^4} = \frac{0.2048}{0.4096} = \frac{1}{2}$.$\frac{2p}{q} = \frac{1}{2} \Rightarrow 4p = q$.Since $q = 1 - p$,we have $4p = 1 - p \Rightarrow 5p = 1 \Rightarrow p = \frac{1}{5}$.Then $q = 1 - \frac{1}{5} = \frac{4}{5}$.Now,the probability of exactly $3$ successes is:$P(X=3) = {}^5C_3 p^3 q^2 = 10 	imes (\frac{1}{5})^3 	imes (\frac{4}{5})^2$.$P(X=3) = 10 	imes \frac{1}{125} 	imes \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}$.

Let in a Binomial distribution,consisting of $5$ independent trials,probabilities of exactly $1$ and $2$ successes be $0.4096$ and $0.2048$ respectively. Then the probability of getting exactly $3$ successes is equal to

Similar Questions

$A$ die is rolled $5$ times. Getting an odd number in one trial is considered as a success. The variance of the distribution of successes is

$A$ fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to that of getting $9$ heads,then the probability of getting $3$ heads is

The probability that a certain kind of component will survive a given test is $\frac{2}{3}$. The probability that at most $2$ components out of $4$ tested will survive is

If a variable takes values $0, 1, 2, \dots, n$ with frequencies ${q^n}, \frac{n}{1}{q^{n - 1}}p, \frac{n(n - 1)}{1 \times 2}{q^{n - 2}}{p^2}, \dots, {p^n}$,where $p + q = 1$,then the mean is

The probability that a student is not a swimmer is $\frac{1}{5}$. Then,the probability that out of $5$ students,$4$ are swimmers is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module