# Blood types A, B, O, and AB have percentages 40%, 20%, 30% and 10%. If two individuals are chosen at random, assuming independence, how do you find probability that both blood type A?

Sep 11, 2016

$0.16$.

#### Explanation:

Let, ${A}_{1}$ be the Event that the randomly chosen First Individual has

the Blood Group $A$, and, let, ${A}_{2}$ be for the Second Individual.

$\therefore \text{ The Reqd. Prob.=} P \left({A}_{1} \cap {A}_{2}\right)$.

But, ${A}_{1} \mathmr{and} {A}_{2}$ are Independent Events, therefore,

$P \left({A}_{1} \cap {A}_{2}\right) = P \left({A}_{1}\right) \cdot P \left({A}_{2}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(\star\right)$.

Since, 40% of Individuals have Blood Group A, clearly,

$P \left({A}_{1}\right) = P \left({A}_{2}\right) = \frac{40}{100} = 0.4$

Hence, by $\left(\star\right) , \text{the Reqd. Prob.} = 0.4 \times 0.4 = 0.16$.