Question

There are 1 0 balls in a bag, 4 red balls, 2 black balls, and 4 yellow balls. Every time you can pick one ball without replacement. Now you pick 2 times.

(1) How likely you will have 2 black ball ls if you pick 2 times.

(2) How likely you will have one red ball and one yellow ball if you pick 2 times.

(3) How likely you will have at least one yellow balls.

Can someone explain it by using nCr (PROB using calculator TI-84) ?

Thanks

Answer 1

1)

P(2 black balls )=P(first black and 2nd black) =(2/10)*(1/9) =1/45 =0.0222

(for ti -84 , use ₂C₂/₁₀C₂ )

2)

P( one red ball and one yellow ball if you pick 2 times) =P(first red and 2nd yellow)+P(1st yellow and 2nd red) =(4/10)*(3/9)+(4/10)*(3/9)=24/90 =4/15 =0.2667

((for ti -84 , use ₄C₁*₄C₁/₁₀C₂ )

3)

P(at least one yellow )=1-P(none of them is yellow) =1-(6/10)*(5/9)=2/3 =0.6667\

((for ti -84 , use 1- ₆C₂/₁₀C₂ )