InterviewSolution
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InterviewSolution
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Three numbers are chosen at random without replacement from {1, 2, 3,...... 8}. The probability that their minimum is 3, given that their maximumis 6, is(1) `3/8`(2) `1/5`(3) `1/4`(4) `2/5` |
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Answer» let event A= minimun is 3 let event B= maximum is 6 total ways = `.^8C_3` `P(A nn B) = 2/(.^8C_3)= 2/((8!)/(3!5!))` `= (2 xx 3!)/(8 xx 7 xx6) = 1/28` `P(A/B) = (P(A nn B))/(P(B))` `= (1/28)/((.^5C_2)/(.^8C_3))` `= (2/(.^8C_3))/((.^5C_2)/(.^8C_3)` `= 2/(.^5C_2)` `= ( 2 xx2)/(5 xx4) = 1/5 ` option 1 is correct |
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