Three coins are tossed together. Find the probability of getting:

(iii) at least one head and one tail

Asked by Aaryan | 1 year ago |  46

Solution :-

Given: Three coins are tossed together.

Number of possible outcomes is 2= 8

(i) Let E be the event of getting exactly two heads

E = {(H, H, T) (H, T, H) (T, H, H)}

n (E) = 3

P (E) =$$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{3} {8}$$

(ii) Let E be the event of getting at least two heads

E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}

n (E)=4

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{4} {8}$$

$$\dfrac {1}{2}$$

(iii) Let E be the event of getting at least one head and one tail

E = {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}

n (E) = 6

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{6} {8}$$

$$\dfrac {3}{4}$$

Answered by Aaryan | 1 year ago

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