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# easy maths

1)A coin is thrown twice.Find the probability that

a)all are heads

b)at least one is head

c)exactly one is tail

Any explaination?

### 2 Answers

- 1 decade agoFavorite Answer
(a)

1st way of thinking:

possible cases: HT, HH, TH, TT

Pr[all heads] = correct case / possible case = 1/4

2nd way of thinking (assume the two outcomes are independent):

Pr[head]=1/2

Pr[both head] = Pr[1st head] x Pr[2nd head] = (1/2)^2 = 1/4

(b)

Pr[at least one head]

= 1-Pr[no head]

= 1-Pr[all tails]

= 1-1/4 = 3/4

(c)

Possible cases: HH, [HT], [TH], TT

Pr[exactly one tail]

=2/4 =1/2

Source(s): myself