Possibility & Combination problem

Question

need help,

if a bag has

5 sweet candy

4 choco candy

 

the possibility to get 1 sweet candy is 5/9

                              2 sweet candy is 5/9 x 4/8

the possibility to get 1 choco candy is 4/9

                              2 choco candy is 4/9 x 3/8

 

then how much is the possibility to get a) 1 sweet candy and 1 choco candy?

                                                        b) 2 sweet candy and 1 choco candy?

 

the formula my teacher teach me kinda struck with my logical way. so i really want to know which is right.

 

please write the solution with details

Answer 1

You have the right idea. 

 

P (1s and 1c) = P(s,c) + P(c,s) = (5/9)(4/8) + (4/9)(5/8) = .555

 

P(2s and 1c) = P(s,s,c) + P(s,c,s) + P(c,s,s) = (5/9)*(4/8)*(4/7) + (5/9)*(4/8)*(4/7) + (4/9)*(5/8)*(4/7) = .476

 

 

An easier way to solve this is using a hypo-geometric distribution, which is a formula for the combinations of a sample divided by the combination of a population.

 

P(1s and 1c) 

= combinations of 1 sweet * combination of 1 choc / combinations of 2 candies

= C(5,1) * C(4,1)/C(9,2) = .555

 

P(2s and 1c)

= combinations of 2 sweets * combinations of 1 choc / combinations of 3 candies

=  C(5,1) * C(4,1)/C(9,3) = .476

 

Tip: A really easy way to calculate these is using the HYPGEOM.DIST function in Excel.