A fair die with faces 1, 2, 3, 4, 5, and 6 is | vedclass

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(A) Let $X$ be the outcome of a single toss of the die.We are given that the value must be $\ge 2$ and $\le 5$.The favorable outcomes are $\{2, 3, 4,

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let $X$ be the outcome of a single toss of the die.We are given that the value must be $\ge 2$ and $\le 5$.The favorable outcomes are $\{2, 3, 4, 5\}$.The number of favorable outcomes is $4$.The total number of outcomes on a die is $6$.The probability of success in a single trial is $p = \frac{4}{6} = \frac{2}{3}$.The probability of failure in a single trial is $q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3}$.Since the die is tossed $4$ times,we use the binomial distribution formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$ where $n=4$ and $k=4$.The required probability is $P(X = 4) = \binom{4}{4} \left(\frac{2}{3}ight)^4 \left(\frac{1}{3}ight)^0 = 1 	imes \frac{16}{81} 	imes 1 = \frac{16}{81}$.

$A$ fair die with faces $1, 2, 3, 4, 5,$ and $6$ is tossed $4$ times. What is the probability that in all $4$ tosses,the value obtained is not less than $2$ and not greater than $5$ (in $/81$)?

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