Question

The bucket containing orange tennis balls and yellow tennis balls from which 5 balls are selected at random, but assume that the bucket contains 9 orange balls and 7 yellow balls. (1) what is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow? (1) what is the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow?

Answer 1

(1)  

Probability that all are yellow = $\frac{7\times6\times5\times4\times3}{16\times15\times14\times13\times12}=0.0048$

Probability that all are orange = $\frac{9\times8\times7\times6\times5}{16\times15\times14\times13\times12}=0.0288$

Sum of the two probabilities = 0048+ 0.0288=0.0336

Thus, the probability that at least one is orange and at least one is yellow = 1-0.0336= 0.9664= 96.64 %

(2) In addition to the above, we eliminate the "1 and 4" cases.  

In these cases, any of the 5 can be taken out, so we multiply by 5.  

1 yellow + 4 orange = $5\times \frac{(7 \times 9 \times 8 \times 7 \times 6)}{(16 \times 15 \times 14 \times 13 \times12)} = 0.2019$

1 orange + 4 yellow = $5 \times \frac{(9 \times 7\times 6 \times 5 \times 4)}{(16 \times 15\times 14 \times 13 \times 12)} = 0.0721$

Sum of two probabilities = 0.2019+0.0721=0.2740  

Now adding 0.0336 to this probability =0.0336+0.2740= 0.3076  

Subtracting that from 1.0000 = 1-0.3076=0.6924= 69.24 %

Thus, the probability that, of the 5 balls selected at random, at least two are orange and at least two are yellow= 69.24%