A person fails 4 times in a game when he play | vedclass

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

Question

(C) Given that the person fails $4$ times in $9$ games, the number of successes is $9 - 4 = 5$. Thus, the probability of success in a single game is $

Vedclass · Accepted Answer

$\frac{79}{9}\left(\frac{4}{9}\right)^{14}$

Vedclass · Answer

(C) Given that the person fails $4$ times in $9$ games, the number of successes is $9 - 4 = 5$. Thus, the probability of success in a single game is $p = \frac{5}{9}$. Consequently, the probability of failure is $q = 1 - p = 1 - \frac{5}{9} = \frac{4}{9}$. For $n = 15$ trials, we want the probability of at most one success, i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$. Using the binomial distribution formula $P(X = r) = {}^{n}C_{r} p^r q^{n-r}$: $P(X = 0) = {}^{15}C_{0} \left(\frac{5}{9}ight)^0 \left(\frac{4}{9}ight)^{15} = \left(\frac{4}{9}ight)^{15}$. $P(X = 1) = {}^{15}C_{1} \left(\frac{5}{9}ight)^1 \left(\frac{4}{9}ight)^{14} = 15 	imes \frac{5}{9} 	imes \left(\frac{4}{9}ight)^{14} = \frac{75}{9} 	imes \left(\frac{4}{9}ight)^{14}$. Adding these probabilities: $P(X \leq 1) = \left(\frac{4}{9}ight)^{15} + \frac{75}{9} 	imes \left(\frac{4}{9}ight)^{14} = \left(\frac{4}{9}ight)^{14} \left[ \frac{4}{9} + \frac{75}{9} ight] = \frac{79}{9} \left(\frac{4}{9}ight)^{14}$.

$A$ person fails $4$ times in a game when he plays $9$ times. If he plays $15$ times, the probability of having success at most one is

Similar Questions

Six balls are drawn successively from an urn containing $7$ red and $9$ black balls. Determine whether or not the trials of drawing balls are Bernoulli trials when after each draw,the ball drawn is replaced.

Minimum number of times a fair coin must be tossed,so that the probability of getting at least one head,is more than $99 \%$ is

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. Then,the probability that $X$ takes values greater than $1$ is

The random variable $X$ has a Binomial distribution $B(20, 0.4)$. Then $5 - 5 P(X \geq 2) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module