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Balachandar S
This is which year question paper
It's a practice book of MCC that was provided me by my friend. Who passed in previous attempt.
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Virendra SainiQuestion no 40 and 41.
Q 40 - answer will be b) Take the probability of defective item as 0.03 (3%) and non-defective item as 0.97. So, probability that in a sample of 4 items all are non-defective will be 0.97^4 = 0.885 Q 41 - answer will be a) Total number of ways of distributing 8 balls in 3 boxes is 3^8 = 6561 (i.e,, balls can go in any of 3 boxes, so total ways become 3 x 3 x 3 x .. 8 times) Ways in which first box can have 3 balls is 8C3 x 2^5 = 1792 (i.e. any 3 balls chosen out of 8, remaining 5 balls can go in any of the 2 box - 2 x 2 x 2 x... 5 times) So, probability becomes 1792/6561
Sahibdeep Singh
Q 40 - answer will be b) Take the probability of defective item as 0.03 (3%) and non-defective item as 0.97. So, probability that in a sample of 4 items all are non-defective will be 0.97^4 = 0.885 Q 41 - answer will be a) Total number of ways of distributing 8 balls in 3 boxes is 3^8 = 6561 (i.e,, balls can go in any of 3 boxes, so total ways become 3 x 3 x 3 x .. 8 times) Ways in which first box can have 3 balls is 8C3 x 2^5 = 1792 (i.e. any 3 balls chosen out of 8, remaining 5 balls can go in any of the 2 box - 2 x 2 x 2 x... 5 times) So, probability becomes 1792/6561
Way of understanding is so simple I understand the answer in seconds, where that I didn't understand it on practice of entire chapter. I appreciate it.