If on an average 9 mountain climbers out of 1 | vedclass

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(B) Let $p$ be the probability of a climber returning safely,so $p = \frac{9}{10}$.Let $q$ be the probability of a climber not returning safely,so $q

Vedclass · Accepted Answer

$\frac{9^4 	imes 7}{50000}$

Vedclass · Answer

(B) Let $p$ be the probability of a climber returning safely,so $p = \frac{9}{10}$.Let $q$ be the probability of a climber not returning safely,so $q = 1 - p = \frac{1}{10}$.We use the binomial distribution formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$,where $n = 5$.We need to find the probability that at least $4$ climbers return safely,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.$P(X = 4) = {}^5C_4 \cdot (\frac{9}{10})^4 \cdot (\frac{1}{10})^1 = 5 \cdot \frac{9^4}{10^5}$.$P(X = 5) = {}^5C_5 \cdot (\frac{9}{10})^5 = 1 \cdot \frac{9^5}{10^5}$.$P(X \ge 4) = \frac{5 \cdot 9^4}{10^5} + \frac{9^5}{10^5} = \frac{9^4(5 + 9)}{10^5} = \frac{9^4 \cdot 14}{100000} = \frac{9^4 \cdot 7}{50000}$.

If on an average $9$ mountain climbers out of $10$ return safely,what is the probability that with $5$ climbers out,at least $4$ will return safely?

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