Question

if the three coins are simultaneously tossed again compute the probability of getting more heads than Tails and getting more tails than head

Answer 1

{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. We see that {HHH, HHT, HTH, THH} have at least two heads. Therefore 4 out of 8 outcomes have at least two heads. P(at least 2 heads) = 4/8 = 1/2.
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Answer 2

HI mate here is your answer..,

Three coins are tossed simultaneously

:. the sample space

S={HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

:.n(S)=8

Event A getting more head than tail

Then A={HHH, HHT, HTH, THH}

:. n(A)=4

p(A)=n(A)/n(S)

=4/8
=1/2

Event B getting more tail than head

Then B={HTT, TTH, THT, TTT}

:.n(B)=4

p(B)=n(B)/n(S)
=4/8
=1/2

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