A strategy makes 8 independent trades, and each trade wins with probability 1/2. What is the probability that it wins at least 6 of them?

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Winning most of the trades

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A strategy makes $8$ independent trades, and each trade wins with probability $\frac{1}{2}$ . What is the probability that it wins at least $6$ of them?

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Hint

Sum the binomial coefficients for $6$ , $7$ and $8$ wins.

Solution

The total number of outcomes is $2^{8} = 256$ .
Exactly $6$ wins can occur in $28$ ways.
Exactly $7$ wins can occur in $8$ ways, and exactly $8$ wins in $1$ way.
The favorable count is $28 + 8 + 1 = 37$ .
The probability is $\frac{37}{256}$ .

Answer: $\frac{37}{256}$

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