91Ó°ÊÓ

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

The proportions (probabilities) of blood phenotypes in the US population are supplied to us. Assume we choose two people at random; the chance that both phenotypes are O is

\(\begin{array}{c}P(O \cap O) &=& P(O) \cdot P(O)...........(1)\\ &=& 0.45 \cdot 0.45\\ &=& 2025\end{array}\)

(1): employing the below-mentioned multiplication property (we are given that selections are independent of one another).

Two occurrences A and B are independent if and only if they have the same multiplication property.

\(P(A \cap B) = P(A) \cdot P(B)\)

We are also asked to find the probability that the phenotypes of two randomly selected individuals match, which can be represented as union of four disjoint events\(,B \cap B,(AB) \cap (AB)\), and \(A \cap A\)

\(\begin{array}{c}P[(O \cap O) \cup (B \cap B) \cup ((AB) \cap (AB)) \cup (A \cap A)]\\ &=& P(O \cap O) + P(B \cap B) + P(AB \cap AB) + P(A \cap A)...............(1)\\ &=& P(O) \cdot P(O) + P(B) \cdot P(B) + P(AB) \cdot P(AB) + P(A) \cdot P(A)...............(2)\\ &=& 0.45 \cdot 0.45 + 0.11 \cdot 0.11 + 0.04 \cdot 0.04 + 0.4 \cdot 0.4\\ &=& 0.2025 + 0.0121 + 0.0016 + 0.16\\ &=& 0.3762,\end{array}\)

(1): the events are disjoint;

(2): the multiplication property.

Therefore, the result is

\(\begin{array}{c}P(O \cap O) &=& 0.2025;\\P((O \cap O) \cup (B \cap B) \cup ((AB) \cap (AB)) \cup (A \cap A)) &=& 0.3762\end{array}\)