A fair coin is tossed $4$ times. The probability that heads exceed tails in number is?

\frac{3}{16}

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\frac{1}{4}

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\frac{5}{16}

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\frac{7}{16}

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Solution

The correct option is C $\frac{5}{16}$
Total number of outcomes=222*2=16
Events: HHHH p1=1/16
HHHT= $(1 / 16) \times (4! / 3!)$ =1/4
Total probability=5/16

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A fair coin is tossed 4 times. The probability that heads exceed tails in number is?

The correct option is C 516Total number of outcomes=2*2*2*2=16Events: HHHH p1=1/16 HHHT=(1/16)×(4!/3!)=1/4Total probability=5/16

A fair coin is tossed $4$ times. The probability that heads exceed tails in number is?

The correct option is C $\frac{5}{16}$
Total number of outcomes=222*2=16
Events: HHHH p1=1/16
HHHT= $(1 / 16) \times (4! / 3!)$ =1/4
Total probability=5/16