Quality assurance-TVs inspected in lots of 100, 5 of the 100 selected at random.

Question

Quality assurance-TVs inspected in lots of 100, 5 of the 100 selected at random.
Assuming that 6 of the 100 TVs are defective, find the probability that exactly 1 of the 5 TVs selected are defective.
and at least one is defective
and at most one is defective

Thanks

Explanation / Answer

A={TV is defective} Z={not defective} P(A)=6%=0.06 The probability that exactly 1 of 5 TVs is defective is the probability that 4 are not defective times the probability that one is defective times the number of positions the defective TV can be in. The following list shows the possible positions. AZZZZ ZAZZZ ZZAZZ ZZZAZ ZZZZA (the first row means that the first TV is defective, the next four are not. The second row means that the second TV is defective, etc.) --> P(1 in 5 TVs is defective)=0.94^4*0.06*5=0.234 The probability that at least one is defective equals 1 minus the probability that none is defective P(none is defective)=0.94^5=0.734 --> P(at least one is defective)=1-0.734=0.266 The probability that at most one is defective equals the probability that none is defective plus the probability that one is defective --> P(at most one is defective)=0.734+0.234=0.968

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