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Misc 10 Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in color. Find the probability that the transferred ball is black. Let A : Event of drawing red ball from Bag II E1 : Event that red ball is transferred from Bag I E2 : Event that black ball is transferred from Bag I We need to find out the probability that the ball transferred is black, if ball drawn is red in color i.e. P(E2|A) P(E2|A) = (𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2) ) "P(E1)" = Probability that red ball is transferred from Bag I = 𝟑/𝟕 P(A|E1) = Probability that red ball is drawn from bag II ,if red ball is transferred from Bag I = 5/10 = 𝟏/𝟐 "P(E2)" = Probability that black ball is transferred from Bag I = 𝟒/𝟕 P(A|E2) = Probability that red ball is drawn from bag II ,if black ball is transferred from Bag I = 4/10 = 𝟐/𝟓 Now, P(E2|A) = (𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2))/(𝑃(𝐸_1 ).𝑃(𝐴|𝐸_1)+𝑃(𝐸_2 ).𝑃(𝐴|𝐸_2) ) = (4/7 ×2/5 )/(3/7 ×1/2 + 4/7 ×2/5) = (8/35 )/(3/14+ 8/35) = (8/35 )/((15+16)/70 ) = (𝟏𝟔 )/𝟑𝟏 = (8/35 )/(3/14+ 8/35) = (8/35 )/((15+16)/70 ) = (𝟏𝟔 )/𝟑𝟏

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo