HW5_sol

HW5_sol - 0; if = 1 / 4, the most likely number of heads is...

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CSE21 FA11 Homework #5 (10/17/11) 5.1 Solution. (a) The probability that no Ace appears on any of the draws is ( 48 52 ) 10 . The probability that at least one King appears in 10 draws is 1 - ( 48 52 ) 10 . The probability that at least 2 Queens appear in the 10 draws is 1 - ( 48 52 ) 10 - 48 9 × ( 10 1 )( 4 1 ) 52 10 . (b) The probability that no Ace appears on any of the draws is ( 48 10 ) / ( 52 10 ) . The probability that at least one King appears in 10 draws is 1 - ( 48 10 ) / ( 52 10 ) . The probability that at least 2 Queens appear in the 10 draws is 1 - ( 48 10 ) ( 52 10 ) - ( 48 9 )( 4 1 ) ( 52 10 ) . 5.2 Solution. Pr ( A B ) = Pr ( A ) + Pr ( B ) - Pr ( A B ) = Pr ( A ) + Pr ( B ) - (1 - Pr ( A B ) c ) = 3 / 7 + 1 / 2 - 5 / 8 = 17 / 56 . 5.3 Solution. The probability that there is no head is 1 2 10 . The probability that there is just one head is ( 10 1 ) 2 10 . The probability that there are two heads is ( 10 2 ) 2 10 . Thus, the probability that there are at most two heads is 1+10+45 2 10 = 56 1024 . 5.4 Hints. The probability that there are i ( i = 0 , 1 , 2 , 3) heads is ( 3 i ) α i (1 - α ) 3 - i . It is easy to verify that if α [0 , 1 / 4), the most likely number of heads is
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Unformatted text preview: 0; if = 1 / 4, the most likely number of heads is 0 or 1; if (1 / 4 , 1 / 2), 1 the most likely number of heads is 1; if = 1 / 2, the most likely number of heads is 1 or 2; if (1 / 2 , 3 / 4), the most likely number of heads is 2; if = 3 / 4, the most likely number of heads it 2 or 3; if (3 / 4 , 1], the most likely number of heads it 3. 5.5 Solution. 12 identical jellybeans are distributed randomly to 5 students. The probability that each student gets at least one jellybean is ( 12-5+5-1 5-1 ) ( 12+5-1 5-1 ) . The probability that each student gets at least two jellybean is ( 12-5 2+5-1 5-1 ) ( 12+5-1 5-1 ) . 5.6 Solution. The probability in the general case that none of 10 DVDs you nally end up with are defective is ( 70+ k k ) ( 90 k ) . 2...
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HW5_sol - 0; if = 1 / 4, the most likely number of heads is...

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