A die is thrown 6 times. If 'getting an odd n | vedclass

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(A) The repeated tosses of a die are Bernoulli trials. Let $X$ denote the number of successes of getting odd numbers in an experiment of $6$ trials. P

Vedclass · Accepted Answer

$\frac{63}{64}$

Vedclass · Answer

(A) The repeated tosses of a die are Bernoulli trials. Let $X$ denote the number of successes of getting odd numbers in an experiment of $6$ trials. Probability of getting an odd number in a single throw of a die is $p = \frac{3}{6} = \frac{1}{2}$. Therefore,$q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$. $X$ follows a binomial distribution $B(n, p)$ with $n = 6$ and $p = \frac{1}{2}$. The probability mass function is given by $P(X = x) = ^{6}C_{x} \left(\frac{1}{2}\right)^{6-x} \left(\frac{1}{2}\right)^{x} = ^{6}C_{x} \left(\frac{1}{2}\right)^{6}$. We need to find the probability of at most $5$ successes,which is $P(X \leq 5)$. Using the complement rule,$P(X \leq 5) = 1 - P(X > 5) = 1 - P(X = 6)$. $P(X = 6) = ^{6}C_{6} \left(\frac{1}{2}\right)^{6} = 1 \cdot \frac{1}{64} = \frac{1}{64}$. Therefore,$P(X \leq 5) = 1 - \frac{1}{64} = \frac{63}{64}$.

$A$ die is thrown $6$ times. If 'getting an odd number' is a success,what is the probability of at most $5$ successes?

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