Three coins are tossed simultaneously.find the probability of getting atleast two heads.
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Answered by
11
Total number of coins = 3.
Total number of possible outcomes n(S) = 2^3 = 8.
Let A be the event of getting at least two heads.
n(A) = {HHH,HTH,HHT,THH}
= 4.
Therefore, required probability P(A) = n(A)/n(S)
= > 4/8.
= > 1/2.
Hope this helps!
siddhartharao77:
:-)
Answered by
3
three times double head can come
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