Question

We are flipping a coin 6 times. What is the probability of the outcomes to be :

(Binomial distribution).

a) two tails.

b) At least 5 tails.

c) The most 2 heads

Answer 1

Solution:

$p=q=\frac12$

$n=6$

$X\sim Bin(n,p)$

a) P(two tails)=P(X=2)$=^6C_2(0.5)^2(0.5)^4=0.234375$

b) At least 5 tails)=$P(X\ge5)=P(X=5)+P(X=6)$

$=^6C_5(0.5)^5(0.5)^1+^6C_6(0.5)^6(0.5)^0 \\=6(0.5)^6+(0.5)^6 \\=0.109375$

c) P(The most 2 heads)$=P(X\le2)=P(X=0)+P(X=1)+P(X=2)$

$=^6C_0(0.5)^0(0.5)^6+^6C_1(0.5)^1(0.5)^5+^6C_2(0.5)^2(0.5)^4 \\=(0.5)^6+6(0.5)^6+15(0.5)^6 \\=0.34375$