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A grocer's bin contains 7 ripe tomatoes and 9 unripened tomatoes.
Three tomatoes are selected at random (Tomatoes are not replaced).
Show the probability distribution for the number of ripe tomatoes.
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Let x be the number of ripe tomatoes in our random sample of 3 tomatoes.
They want you find and present the numbers P(x=0), P(x=1), P(x=2) and P(x=3).
(0) x=0 means that in our sample all 3 tomatoes are unripped; there is no ripped tomatoes.
So, P(x=0) = = = .
Here = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.
= 84 in the numerator is the number of triples of 9 unripped tomatoes.
(1) x = 1 means that in our sample 1 tomato is ripped; two other tomatoes are unripped.
So, P(x=1) = = = = .
Again, here = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.
= 36*7 = 252 in the numerator is the number of triples,
consisting of 2 unripped tomatoes and 1 ripped tomato.
(2) x = 2 means that in our sample 2 tomatoes are ripped; one tomato is unripped.
So, P(x=2) = = = = .
Again, here = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.
= 9*21 = 189 in the numerator is the number of triples,
consisting of 1 unripped tomato and 2 ripped tomatoes.
(3) x = 3 means that in our sample 3 tomatoes are ripped; no unripped tomatoes.
So, P(x=3) = = = = after reducing.
Again, here = 560 in the denominator is the total number of triples of 16 = 7+9 tomatoes.
= 1*35 = 35 in the denominator is the number of triples,
consisting of 0 unripped tomatoes and 3 ripped tomatoes.
Thus P(0) = ; P(1) = ; P(2) = ; P(3) = = .
The sum P(0) + P(1) + P(2) + P(3) = + + + = + = + = = 1.
Solved.