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

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(C) Given,$n=10$. Probability of getting an odd number in a single throw,$p = \frac{3}{6} = \frac{1}{2}$. Probability of not getting an odd number,$q

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$\frac{1023}{1024}$

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(C) Given,$n=10$. Probability of getting an odd number in a single throw,$p = \frac{3}{6} = \frac{1}{2}$. Probability of not getting an odd number,$q = 1 - p = \frac{1}{2}$. We need to find the probability that an odd number appears at least once,which is $P(X \geq 1)$. Using the complement rule,$P(X \geq 1) = 1 - P(X=0)$. Using the binomial distribution formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$,we have: $P(X=0) = {}^{10}C_{0} (\frac{1}{2})^{0} (\frac{1}{2})^{10} = 1 	imes 1 	imes \frac{1}{1024} = \frac{1}{1024}$. 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 once is

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