Question

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?

Answer 1

`= 7/44`

Explanation:

For:

• `12 = {(5R),(4G ),(3Y ):}`

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 = k) = (((K),(k)) ((N-K),(n-k)))/(((N),(n)))`

• `N = 12`, is the population size (ie, # balls):

• `K= 5`, is the number of success states (ie, #Red balls) in the population,
• `n= 3`, is the number of draws
• `k= 0`, is the number of observed successes

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

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