A box has 6 black, 4 red, 2 white and 3 blue shirts. What is the probability that 2 red shirts and 1 blue shirt get chosen during a random selection of 3 shirts from the box?
Correct Answer: $$\frac{{18}}{{455}}$$
We want 2 red and 1 blue shirt
There are 4 red shirts and 3 blue shirts
Total = 15 shirts
You can choose blue shirt 1st, then red shirt
And then red shirt Probability
$$\eqalign{ & = \frac{3}{{15}} \times \frac{4}{{14}} \times \frac{3}{{13}} \cr & = \frac{6}{{455}} \cr} $$
Or you can choose red shirt 1st, then red shirt and then blue shirt
Or you can choose red shirt 1st, then blue shirt and then red shirt
For all 3 the probability remains same = $$\frac{6}{{455}}$$
We need to add these 3 probabilities to get total probability
∴ Total probability
$$\eqalign{ & = \frac{6}{{455}} + \frac{6}{{455}} + \frac{6}{{455}} \cr & = \frac{{18}}{{455}} \cr} $$
There are 4 red shirts and 3 blue shirts
Total = 15 shirts
You can choose blue shirt 1st, then red shirt
And then red shirt Probability
$$\eqalign{ & = \frac{3}{{15}} \times \frac{4}{{14}} \times \frac{3}{{13}} \cr & = \frac{6}{{455}} \cr} $$
Or you can choose red shirt 1st, then red shirt and then blue shirt
Or you can choose red shirt 1st, then blue shirt and then red shirt
For all 3 the probability remains same = $$\frac{6}{{455}}$$
We need to add these 3 probabilities to get total probability
∴ Total probability
$$\eqalign{ & = \frac{6}{{455}} + \frac{6}{{455}} + \frac{6}{{455}} \cr & = \frac{{18}}{{455}} \cr} $$