Question

A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads, and (ii) at most 4  heads. 

Answer 1

Total number of the probability of tossing a coin 5 times is 32

(i) Probability of getting at least 4 heads 

P(X=4) + P(X=5)

`""^5C_4 (1/2)^1 (1/2)^4 + ""^5C_5 (1/2)^0 (1/5)^5`

= `""^5C_4 (1/2)^5 + ""^5C_5 (1/2)^5`

= `6/32 = 3/16`

(ii) Probability of getting at most 4 head

P(X=1) + P(X=2) + P(X=3) + P(X=4)

`""^5C_1 (1/2)^5 + ""^5C_2 (1/2)^5 + ""^5C_3 (1/2)^5 + ""^5C_4 (1/2)^5`

= `(1/2)^5` [5+ 10 + 10 + 5]

= `(15)/(16)`