The probability that there will be 0 defective sets:
p(0)=C73C53=3!2!5!⋅7!3!4!=5⋅6⋅73⋅4⋅5=72
1 defective sets:
p(1)=C73C52C21=2⋅3!2!5!⋅7!3!4!=2⋅5⋅6⋅73⋅4⋅5=74
2 defective sets:
p(2)=C73C51C22=5⋅7!3!4!=5⋅5⋅6⋅76=71
We have the probability distribution:
Xp072174271
Probability histogram:
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