A bag contains 12 sweets, of which 5 are red, 4 are green and 3 are yellow. 3 sweets are chosen without replacement. What is the probability that he chooses no red sweets?

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Answer 1 · 2018-06-30T23:55:28

$= 7/44$

For:

There are $((12),(3))$ ways of choosing 3 sweets from 12, regardless of color

There are $((7),(3))$ ways of choosing 3 sweets from the non-red sweets.

So:

$P("all 3 are non-Red") = (((7),(3)))/(((12),(3))) = (7!9!3!)/(3!4!12!)$

$= (7*6*5 )/( 12*11*10) = 7/44$

Alternatively, using the Hypergeometric distribution:

$P(X = 0) = (((12),(0)) ((7),(3)))/(((12),(3)))$

$= (1* ((7!)/(3!4!)))/(((12!)/(9!3!))) = 7/44$