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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?

A. $$\frac{{18}}{{455}}$$

B. $$\frac{{7}}{{15}}$$

C. $$\frac{{7}}{{435}}$$

D. $$\frac{{7}}{{2730}}$$

Answer: Option A

Solution(By Examveda Team)

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
$$\eqalign{ & = \frac{6}{{455}} + \frac{6}{{455}} + \frac{6}{{455}} \cr & = \frac{{18}}{{455}} \cr} $$

This Question Belongs to Arithmetic Ability >> Probability

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Comments ( 1 )

  1. Naseer Muhammad
    Naseer Muhammad :
    10 months ago

    We can easily solve this as
    4C2×3C1
    ----------- = 18/455 . Same result
    15C3

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