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?

1 Answer
Jun 30, 2018

Answer:

#= 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#