A die is thrown 7 times. The chance that an odd number turns up at least 4 times, is

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$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{6}$

answer is B.

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Detailed Solution

Required probability =

$7_{C_{4}} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{3} + 7_{C_{5}} {(\frac{1}{2})}^{5} {(\frac{1}{2})}^{2} + 7_{C_{6}} {(\frac{1}{2})}^{6} {(\frac{1}{2})}^{1} + 7_{C_{7}} {(\frac{1}{2})}^{7} = \frac{1}{2}$

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