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Supplementary_Questions2_Exams

# Find the probability that two of the selected balls

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Find the probability that two of the selected balls are white, whereas the other one is red. b) Next, assume Box-B contains 5 black similar balls with the same size of the others in Box-A. However, each black ball has a number from 1 to 5 on itself. Assume 2 balls are selected from Box-B at random with no replacement. Find the probability that the sum of the two numbers on these 2 black balls is equal to 6. c) Finally, assume that all the 5 black balls of Box-B is added into Box-A. Then, 5 balls are drawn at random from Box-A without any specific order with no replacement. Find the probability that the selection will yield 2 white, 1 red and 2 black balls, for which sum of the two numbers on these 2 black balls is equal to 6. a) 40 7 3 10 1 7 2 3 b) There are two pairs, (1,5) and (2,4) (with no order), whose sum is equal to 6. Hence, 10 2 2 5 2 . c) P( (“ 2w ”) & ( 1r & 2b with sum 6” ) ) = P(( “2w”) , (“ 1r & 2 b with sum 6” ) | “3 indist. & 2 dist.”) P(“3 indist. & 2 dist.”)= P(“ part (a) “) P (“ part (b) “) P(“3 indist. & 2 dist.”) = 40 7 10 2 143 2 5 15 2 5 3 10

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Q.3(20 pts) There are 5 boxes numbered from 0 to 4. n th box ( n = 0,1,. ..,4) contains n red and (4- n ) black balls (e.g. 3 rd box contains 3 red and 1 black balls). A random number, k , is generated between 0 and 4 with equal probability. Then, a ball is drawn from the k th box. a) What is the probability that the ball is red? b) If the ball is red, what is the probability that the random number equals to “ 3 ” ? c) If the ball is red and a second ball is drawn from the k th box after returning the first one back to the same box. What is the probability that it is red again? a) 2 1 4 4 4 3 4 2 4 1 0 5 1 } { } | { } { 4 0 i i P i r P r P b) 10 3 2 1 5 1 4 3 } { } 3 { } 3 | { } | 3 { r P P r P r P c) 4 3 2 1 16 16 16 9 16 4 16 1 0 5 1 } { } { } | ) {( } { } { } | { 1 4 0 1 2 1 1 2 1 r P i P i r r P r P r r P r r P i r
Q.4 (20 pts) A biased die is rolled 10 times. P{1} = P{2} = P{3} = P{4}=1/6, P{5}=1/12, P{6}=1/4. Find, a) Probability of receiving an even number in five trials. b) Probability of receiving two “4”s and three “6”s . c) Probability of receiving five “6”s, given that five even numbers are observed. a) (Bernouli trials) P(even) = P(2) + P(4) + P(6) = 7 12 P(odd) = 5 12 P( 5 even numbers in 10 trials) =     55 10 10! 7 5 P(even) P(odd) 5 5! 5! 12 12          b) (Multinomial probability law) P(4) = 1 6 P(6) = 1 4 P(other) = P({1,2,3,5}) = 7 12 P( two”4”s and three”6”s) = P( two”4”s, three ”6”s and five “other”s) = 2 3 5 10! 1 1 7 2! 3! 5! 6 4 12                c) (Conditional probability) P(five “6”s | five even and five odd numbers observed) = P((five even and five odd) and (five "6")) P(five even and five odd) 10 15 5 4

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Find the probability that two of the selected balls are...

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