Question

A probability experiment consists of flipping 4 biased coins that land heads only 25% of the time. Let X be the random variable that counts    the number of coins that land heads.

1. Complete the probability distribution for X below.

Distribution of X
X	P (X = k)	P (X ≤ k)
0
1	0.4219
2
3	0.0469
4	0.0039

a.Compute: (i) P (X ≤ 3) and (ii) P (X < 3)

b.Compute: P (X > 1)

c.Compute: P (X is odd)

d.Compute the expected value and variance of X.

Answer 1

Solution

This is case of binomial distribution

For P(X =0) > 4C0*0.25^0*0.75^4 = 0.3164

Similary for P(X = 2) = 4C2*0.25^2*0.75^2 = 0.2109

Completing the table we get

=>

a)

P(X<=3) = 0.9961

P(X<3) = P(X<=2) = 0.9492

b)

P(X>1) = 1- P(X<=1)

= 1-P(X=0) - P(X=1)

= 1-0.3164-0.4219

= 0.2617

C)

p( x = ODD) = P( X = 1) + P(X =3)

= 0.4219+0.0469

= 0.4688

D)

Expected mean:

μ = (0.3164×0+0.4219×1+0.2109×2+0.0469×3+0.0039×4) / (0.3164+0.4219+0.2109+0.0469+0.0039) = 1

Expected variance =

σ² = (0.3164×(0-1)²+0.4219×(1-1)²+0.2109×(2-1)²+0.0469×(3-1)²+0.0039×(4-1)²) / (0.3164+0.4219+0.2109+0.0469+0.0039) = 0.75