Given that A is twice as likely to be selected as B
i.e. P(A) = 2 P(B) …(i)
and C is twice as likely to be selected as D
i.e. P(C) = 2P(D) …(ii)
Now, B and C are given about the same chance
∴ P(B) = P(C) …(iii)
Since, sum of all probabilities = 1
∴ P(A) + P(B) + P(C) + P(D) = 1
⇒ P(A) + P(B) + P(B) + P(D) = 1 [from (iii)]
(a) P(C will be selected) = P(C)
= P(B) [from(iii)]
(b) P(A will not be selected) = P(A’)
= 1 – P(A)
[By complement Rule]