Answer to Question #255686 in Statistics and Probability for Fahim

Question

A fair coin is tossed 5 times. Find the probability of obtaining (i) exactly 4 heads (ii) fewer than 3 heads. Suppose the experiment is a binomial one

Answer 1

The probabilities of falling heads and tails are respectively

$p = q = 0.5$

i) Using Bernoulli's formula, we get:

${P_5}(4) = C_5^4{p^4}q = 5 \cdot {0.5^5} = {\rm{0}}{\rm{.15625}}$

Answer: ${P_5}(4) = {\rm{0}}{\rm{.15625}}$

ii) Let's find the probabilities that there will be 0, 1 and 2 tails:

${P_5}(0) = {q^5} = {0.5^5} = {\rm{0}}{\rm{.03125}}$

${P_5}(1) = C_5^1p{q^4} = 5 \cdot {0.5^5} = {\rm{0}}{\rm{.15625}}$

${P_5}(2) = C_5^2{p^2}{q^3} = 10 \cdot {0.5^5} = {\rm{0}}{\rm{.3125}}$

Then the wanted probability is

$P(x < 3) = {P_5}(0) + {P_5}(1) + {P_5}(2) = 0.03125 + 0.15625 + 0.3125 = 0.5$

Answer: $P(x < 3) = 0.5$