Example 1 : <\/span> 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-<\/p>\n Solution : <\/span>n(S) = \\(^{10}C_4\\) = 210 Example 2 : <\/span> 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.<\/p>\n Solution : <\/span>\\(E_1\\) : Event that first drawn ball is red, second is blue and so on. Example 3 : <\/span> 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.<\/p>\n Solution : <\/span>Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2 Practice these given probability examples to test your knowledge on concepts of probability.<\/p>\n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some probability examples for better understanding of probability concepts. 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 …<\/p>\n
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\\)<\/p>\n
\n\n \n
\\(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\\)\n <\/p>
\n\n \n
\\(\\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<\/p>\n
\n