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 shirtThere 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
$$\eqalign{ & = \frac{6}{{455}} + \frac{6}{{455}} + \frac{6}{{455}} \cr & = \frac{{18}}{{455}} \cr} $$
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Comments ( 1 )
Related Questions on Probability
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
A. $$\frac{{10}}{{21}}$$
B. $$\frac{{11}}{{21}}$$
C. $$\frac{{2}}{{7}}$$
D. $$\frac{{5}}{{7}}$$
A. $$\frac{{1}}{{3}}$$
B. $$\frac{{3}}{{4}}$$
C. $$\frac{{7}}{{19}}$$
D. $$\frac{{8}}{{21}}$$
E. $$\frac{{9}}{{21}}$$
What is the probability of getting a sum 9 from two throws of a dice?
A. $$\frac{{1}}{{6}}$$
B. $$\frac{{1}}{{8}}$$
C. $$\frac{{1}}{{9}}$$
D. $$\frac{{1}}{{12}}$$
We can easily solve this as
4C2×3C1
----------- = 18/455 . Same result
15C3