Question

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? 1 point 18/455 7/15 7/435 7/2730

Answer 1

Total probability of selecting 3 shirts is 15C3

Selection of 2 red shirts can be done in 4C2 = 6

Selection of 1 blue shirt can be done in 3C1 = 3

Therefore probability = 18/15C3 = 18/455

Answer 2

→Total number of shirts in the box = [ 6 black + 4 red + 2 white + 3 Blue ] = 15 shirts

→Probability of an event = $\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}$

→Probability of choosing 2 Red shirts and 1 blue shirt out of 3 shirts chosen

                              =  $\frac{_{2}^{4}\textrm{C}\times_{1}^{3}\textrm{C}}{_{3}^{15}\textrm{C}},\rightarrow _{2}^{4}\textrm{C}= \frac{4!}{2!\times2!}=\frac{3\times4\times2!}{2!\times2}=6, _{1}^{3}\textrm{C} =\frac{3!}{1!\times2!}=3, and _{3}^{15}\textrm{C}=\frac{15!}{12!\times3!}=\frac{13\times14\times15}{3\times2\times1}= 13\times7\times5=455$

            =  $\frac{6\times3}{13\times 7 \times 5} =\frac{18}{455}$

→Option (1) $\frac{18}{455}$ is right answer.