A fair coin is tossed n times such that the p | vedclass

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(B) The probability of getting a head in a single toss is $P(H) = \frac{1}{2}$.The probability of getting no heads in $n$ tosses is $P(X=0) = \left(\f

Vedclass · Accepted Answer

$4$

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(B) The probability of getting a head in a single toss is $P(H) = \frac{1}{2}$.The probability of getting no heads in $n$ tosses is $P(X=0) = \left(\frac{1}{2}\right)^n$.The probability of getting at least one head is $P(X \geq 1) = 1 - P(X=0) = 1 - \left(\frac{1}{2}\right)^n$.Given that $P(X \geq 1) \geq 0.9$,we have:$1 - \left(\frac{1}{2}\right)^n \geq 0.9$Rearranging the inequality:$1 - 0.9 \geq \left(\frac{1}{2}\right)^n$$0.1 \geq \left(\frac{1}{2}\right)^n$$\frac{1}{10} \geq \frac{1}{2^n}$$2^n \geq 10$Testing values for $n$:For $n=3$,$2^3 = 8 For $n=4$,$2^4 = 16 \geq 10$.Thus,the minimum value of $n$ is $4$.

$A$ fair coin is tossed $n$ times such that the probability of getting at least one head is at least $0.9$. Then the minimum value of $n$ is:

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