A and B play a game in which their chances of winning are in the ratio 3:2. The probability of A winning atleast 3 games out of 5 games is

A
0.55
B
0.58
C
0.66
D
0.45

Solution:

$\begin{aligned} p = \frac{3}{5}, q = \frac{2}{5}, n = 5, P (x = r) =^{n} C_{r} {(p)}^{r} {(q)}^{n - r} \\ P (X \geq 3) = P (X = 3) + P (X = 4) + P (X = 5) \\ =^{5} C_{3} {(\frac{3}{5})}^{3} {(\frac{2}{5})}^{2} +^{5} C_{4} {(\frac{3}{5})}^{4} (\frac{2}{5}) +^{5} C_{5} {(\frac{3}{5})}^{5} \end{aligned}$

$= \frac{3^{3}}{5^{5}} (40 + 30 + 9) = \frac{27 \times 79}{3215} = \frac{2133}{3215} = 0.66$

A and B play a game in which their chances of winning are in the ratio 3:2. The probability of A winning atleast 3 games out of 5 games is

Solution:

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