Question

Three unbiased coins are thrown. Find the probability of getting (i) all heads (ii) two heads (iii) 1 head (iv) at least 2 heads (v) at most 2 heads

Answer 1

So total no. of outcomes are 8.
1.so P(E) =1/8
2.so P(E) =3/8 (THH, HHT, HTH)
3.so P(E) =3/8 (TTH, HTT, THT)
4.so P(E) =4/8=1/2(HHT,THH,HHH,HTH)
5.so P(E) =7/8(TTT,HTT,THH,THT,HHT, HTH,HHT)

Answer 2

heya.....

your answer
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total possible outcomes - (HHH, HHT, HTT, TTT, THH, TTH, THT, HTH)

(i) P = 1/8
(ii) P = 3/8
(iii) P = 3/8
(iv) P = 1/2
(v) P = 7/8

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hope it helps ^_^