Question

If a list of 20 individuals who volunteered to supply blood, when it is needed for transfusion, has 15 individuals of type B blood and 3 individuals are selected random from the list, what is the probability that:

a) All 3 will be type B, b) 2 will be of type B and 1 will not, c) at least 1 will be of type B

Answer 1

Answer on Question #72100 – Math – Statistics and Probability

Question

If a list of 20 individuals who volunteered to supply blood, when it is needed for transfusion, has 15 individuals of type B blood and 3 individuals are selected random from the list, what is the probability that:

a) All 3 will be type B,

b) 2 will be of type B and 1 will not,

c) at least 1 will be of type B

Solution

a) $P(3B) = \frac{15C_3}{20C_3} = \frac{455}{1140} = 0.399$;

b) $P(2B) = \frac{15C_2 \times 5C_1}{20C_3} = \frac{105 \times 5}{1140} = 0.461$;

c) $P(\text{at least one B}) = 1 - P(\text{no B})$;

$$P(\text{no B}) = \frac{5C_3}{20C_3} = \frac{10}{1140} = 0.009;$$

$P(\text{at least one B}) = 1 - 0.009 = 0.991$

Answer: a) 0.399; b) 0.461; c) 0.991.

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