Question

Seven fair coins are flipped. The outcomes are assumed to be independent. Let X be the number of heads.

What is the probability that X < 3?

What is the probability that X ≥ 4?

What is the probability that 3 ≤ X < 7

Answer 1

solution:

Given that

No.of trails = No.of fair coins tossing (n) = 7

In the event of tossing a coin

The P(getting head) = P(getting tail ) = 0.5

Let X be the random variable representing No.of heads

Here, X~ B(n,p)  

~B(7,0.5)

P(X=x) = nCx * (p)^x *(1-p)^(n-x)

The probability distribution of X is:

No.of Heads (X)	Probability
0	7C0 * (0.5)^0 *(0.5)^(7) = 0.0078125
1	7C1 * (0.5)^1 *(0.5)^(6) = 0.0546875
2	7C2 * (0.5)^2 *(0.5)^(5) = 0.1640625
3	7C3 * (0.5)^3 *(0.5)^(4) = 0.2734375
4	7C4 * (0.5)^4 *(0.5)^(3) =0.2734375
5	7C5 * (0.5)^5 *(0.5)^(2) = 0.1640625
6	7C6 * (0.5)^6 *(0.5)^(1) = 0.0546875
7	7C7 * (0.5)^7 *(0.5)^(0) = 0.0078125

a) P(X<3)  

P(X<3) = P(X=0) + P(X=1) + P(X=2)

= 0.0078125 + 0.0546875 + 0.1640625

= ~ 0.2266

Therefore, Probability that X<3 =~ 0.2266

b) P(X>=4)

P(X>=4) = P(X = 4) + P(X=5) + P(X=6) + P(X=7)

= 0.2734375 + 0.1640625 + 0.0546875 + 0.0078125

= 0.5000

Therefore, Probability that X>=4 = 0.5000

c) P(3<= X < 7)

P(3<= X < 7) = P(X=3) + P(X =4) + P(X=5) + P(X=6)

= 0.2734375 + 0.2734375 + 0.1640625 + 0.0546875

= ~ 0.7656

Therefore, probability that 3 ≤ X < 7 = 0.7656