Question

Ali draws a ticket from box containing 2 bad, 10 good,and 5 very good tickets.Ahmed draws a next ticket from same box.What is probability of better performance of Ahmed than Ali?

Answer 1

Solution:

Number of bad tickets = 2

Number of good tickets = 10

Number of very good tickets = 5

Total number of tickets = 2+10+5 = 17

P(Ali drawing a bad and Ahmed drawing good ticket or very good ticket)$=\dfrac{^2C_1\times^{10}C_0\times^{5}C_0}{^{17}C_1}\times(\dfrac{^1C_0\times^{10}C_1\times^{5}C_0}{^{16}C_1}+\dfrac{^1C_0\times^{10}C_0\times^{5}C_1}{^{16}C_1})$

$=\dfrac{2}{17}(\dfrac{10}{16}+\dfrac{5}{16})=\dfrac{2}{17}\times \dfrac{15}{16}=\dfrac{15}{136}$

P(Ali drawing a good and Ahmed drawing very good ticket)

$=\dfrac{^2C_0\times^{10}C_1\times^{5}C_0}{^{17}C_1}\times(\dfrac{^2C_0\times^{9}C_0\times^{5}C_1}{^{16}C_1})$

$=\dfrac{10}{17}\times \dfrac{5}{16}=\dfrac{25}{136}$

Now, required probability that Ali performs better than Ahmed$=\dfrac{15}{136}+\dfrac{25}{136}=\dfrac{40}{136}=\dfrac{5}{17}$