Supplementary_Questions2_Exams

# A assume 3 balls are selected as random from this box

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a) Assume 3 balls are selected as random from this box without any specific order with no replacement. 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) =   5 5 5 5 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)

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• Fall '12
• Odtu
• Probability, Probability theory, Anton

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