A die is thrown 10 times. The probability tha | vedclass

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(B) Let $X$ be the number of times an odd number appears in $10$ throws of a die. This follows a binomial distribution with $n = 10$ and $p = \frac{3}

Vedclass · Accepted Answer

$\frac{1023}{1024}$

Vedclass · Answer

(B) Let $X$ be the number of times an odd number appears in $10$ throws of a die. This follows a binomial distribution with $n = 10$ and $p = \frac{3}{6} = \frac{1}{2}$.Thus,$q = 1 - p = \frac{1}{2}$.We need to find the probability of getting an odd number at least once,which is $P(X \geq 1)$.$P(X \geq 1) = 1 - P(X = 0)$.Using the binomial probability formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have:$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{2}ight)^{0} \left(\frac{1}{2}ight)^{10} = 1 	imes 1 	imes \frac{1}{1024} = \frac{1}{1024}$.Therefore,$P(X \geq 1) = 1 - \frac{1}{1024} = \frac{1023}{1024}$.

$A$ die is thrown $10$ times. The probability that an odd number will come up at least one time is:

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