Solution :
We have :
X 0 1 2 3 4
P(X) 0.10 A 0.35 B C
a). P(X < 2) = 0.35
P(X < 2) = P(X = 0) + P(X = 1) = 0.35
⇒ 0.10 + A = 0.35
⇒ A = 0.25
So the value of A is 0.25
b). The total probability = 1
So ,
0.10 + A + 0.35 + B + C = 1
0.10 + 0.25 + 0.35 + B + C = 1
B + C = 1 - 0.70
B + C = 0.30 ......(i)
We have the expected value = 1.9
So, [tex]$\sum X P(X) = 19$[/tex] for x = 0, 1, 2, 3, 4
⇒ (0 x 0.10) + (1 x 0.25) + (2 x 0.35) + (3 x B) + (4 x C) = 1.9
⇒ 0 + 0.25 + 0.70 + 3B + 4C = 1.9
⇒ 3B + 4C = 1.9 - 0.95
⇒ 3B + 4C = 0.95 ...................(ii)
From (i), we take the value of B = 0.30 - C and substitute it in the equation (i), we get,
⇒ 3( 0.30 - C) + 4C = 0.95
⇒ 0.90 - 3C + 4C = 0.95
⇒ C = 0.95 - 0.90
= 0.05
Now substituting the value of C = 0.05 in (ii), we get,
⇒ B = 0.05 = 0.30
⇒ B = 0.25
c). The value of C is 0.05
