Probability Examples

PROBABILITY EXAMPLES

Example 1 : From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers,four persons are selected at random. The probability that selection contains one of each category is-

Solution : n(S) = \(^{10}C_4\) = 210

n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105

\(\therefore\)     P(E) = \(105\over 210\) = \(1\over 2\)



Example 2 : A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

Solution : \(E_1\) : Event that first drawn ball is red, second is blue and so on.

\(E_2\) : Event that first drawn ball is blue, second is red and so on.

\(\therefore\)     P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and

\(\therefore\)     P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\)

P(E) = P(\(E_1\)) + P(\(E_2\)) = 2 \(\times\) \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) = \(6\over 35\)



Example 3 : Three groups A, B, C are contesting for positions on the board of directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group B and C are 0.6 and 0.5 respectively. Find the probability that the new product will be introduced.

Solution : Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2

\(\therefore\) P(A) + P(B) + P(C) = 1

then events A, B, C are exhaustive.

If P(E) = Probability of introducing a new product, then as given

P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5

= 0.5 \(\times\) 0.7 + 0.3 \(\times\) 0.6 + 0.2 \(\times\) 0.5 = 0.35 + 0.18 + 0.10 = 0.63


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