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No.of possible outcomes=8
Possible Outcomes= HHT, HHH, HTH, TTH, TTT, TTH, THT,THH
Atmost two heads means maximum two heads so anything with two heads or less... therefore seven of the possible outcomes have two heads or less. Hence, since no. of possible outcomes is eight and there are seven possibilities with two heads or less the answer is 7/8
Three unbiased coins are totossed. The situation is as follows:
No head : TTT......3C0 = 1 way only this can happen
one head: HTT or THT or TTH = 3C1 = one of these 3ways the event can happen
2heads: HHT or HTH or THH = 3C2 = one of these 3 ways theis event can happen
All 3 heads: HHH onyl 1 way this event,
So in a toss of 3 unbiased coins, the above one of the 4 types of event can happen in any one of 8 different ways. But getting atmost 2heads is the event of 1st type or2nd type or 3rd types in any one of 1+3+3 = 7 ways.
So probablity getting atmost 2 heads = Pr (0 head or 1 head or 2heads ) = pr(0 head)+Pr(1head)+pr(2H) 1/8+3/8+3/8 = 7/8 .
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