In order to get at least one head with a prob | vedclass

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(B) Let $n$ be the number of times the coin is tossed.The probability of getting no heads (all tails) in $n$ tosses is $\left(\frac{1}{2}\right)^n$.Th

Vedclass · Accepted Answer

$4$

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(B) Let $n$ be the number of times the coin is tossed.The probability of getting no heads (all tails) in $n$ tosses is $\left(\frac{1}{2}\right)^n$.The probability of getting at least one head is $1 - \left(\frac{1}{2}\right)^n$.We are given that $1 - \left(\frac{1}{2}\right)^n \ge 0.9$.This implies $\left(\frac{1}{2}\right)^n \le 0.1$.$\Rightarrow \frac{1}{2^n} \le \frac{1}{10}$.$\Rightarrow 2^n \ge 10$.For $n=3$,$2^3 = 8 For $n=4$,$2^4 = 16 \ge 10$.Thus,the minimum number of tosses required is $n = 4$.

In order to get at least one head with a probability $\ge 0.9$,the minimum number of times a coin needs to be tossed is:

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