Answer : Option A

Explanation :

Total number of balls = 4 + 5 + 6 = 15
Let S be the sample space.
n(S) = Total number of ways of drawing 3 balls out of 15 = ¹⁵C₃
Let E = Event of drawing 3 balls, all of them are yellow.
n(E) = Number of ways of drawing 3 balls, all of them are yellow
= Number of ways of drawing 3 balls from the total 5 = ⁵C₃
(∵ there are 5 yellow balls in the total balls)

$MF#%\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}}$MF#%

$MF#% = \dfrac{5_{C_3}}{15_{C_3}} = \dfrac{5_{C_2}}{15_{C_3}}$MF#% [∵ ⁿC_r = ⁿC_{(n - r)}. So ⁵C₃ = ⁵C₂. Applying this for the ease of calculation]

$MF#%= \dfrac{\left(\dfrac{5 \times 4 }{2 \times1}\right)}{\left(\dfrac{15 \times 14 \times 13}{3 \times 2 \times1}\right)}
= \dfrac{5 \times 4}{\left(\dfrac{15 \times 14 \times 13}{3}\right)}
= \dfrac{5 \times 4}{5 \times 14 \times 13} = \dfrac{4}{14 \times 13} = \dfrac{2}{7 \times 13}
= \dfrac{2}{91}$MF#%