Question

It is known that in Canada, the blood types have the following distribution: 46% O, 42% A, 9% B, 3% AB. A randomly chosen Canadian receives a blood transfusion. Knowing that O is a universal donor, A can only donate to A and AB, B can only donate to B and AB, and AB can only donate to AB, what is the probability that the transfusion is not successful? Graph this distribution.

Answer 1

Solution :

It is known that in Canada, the blood types have the following distribution: 46% O, 42% A, 9% B, 3% AB.

Let us say; P(O) = 0.46; P(A) = 0.42; P(B) = 0.09; P(AB) = 0.03.

Let we denote S = succesful transfusion and F = not successful transfusion.

We know that; O is a universal donor, A can only donate to A and AB, B can only donate to B and AB, and AB can only donate to AB.

That is in other words;

If a person with blood group O donates the blood then there is 100% successful transfusion and from this we get P(S | O) = 1 ; P(F | O) = 0.

If a person with blood group A donates the blood then there is (42% + 3% = 45%) successful transfusion whereas (46% + 9%) unsuccessful transfusion and from this we get P(S | A) = 0.45 ; P(F | A) = 0.55.

If a person with blood group B donates the blood then there is (9% + 3% = 12%) successful transfusion whereas (46% + 42%) unsuccessful transfusion and from this we get P(S | B) = 0.12 ; P(F | B) = 0.88.

If a person with blood group AB donates the blood then there is 3% successful transfusion whereas (46% + 42% + 9% = 97%) unsuccessful transfusion and from this we get P(S | AB) = 0.03 ; P(F | AB) = 0.97.

The law of total probability is given by; if A1, A2, A3 be mutually exclusive events and B be any event defined on sample space S then the P(B) = P(A1 and B) + P(A2 and B) + P(A3 and B).

And hence by using this law of total probability we get;

The probability that the transfusion is not successful = P(F) = P(F and O) + P(F and A) + P(F and B) + P(F and AB).

The conditional probability of A given B is; P( A | B ) = P(A and B) / P9B).

From this we get; P( A and B) = P(A|B) * P(B).

So we get;

P(F) = P(F | O)*P(O) + P(F | A)*P(A) + P(F | B)*P(B) + P(F | AB)*P(AB)

P(F) = 0*0.46 + 0.55*0.42 + 0.88*0.09 + 0.97*0.03 = 0 + 0.231 + 0.0792 + 0.0291 = 0.3393.

So required probabiliy that the transfusion is not successful is = 0.3393.