JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 5)
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :
$$\frac{1}{6}$$
$$\frac{2}{15}$$
$$\frac{5}{24}$$
0.08
Explanation
Required probability $=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$
Taking $\mathrm{D}_{(15)}$ as $\frac{15 !}{e}$
$\mathrm{D}_{(14)}$ as $\frac{14 \text { ! }}{e}$
$\mathrm{D}_{(13)}$ as $\frac{13 \text { ! }}{e}$
$$ \text { We get, } \text { Required probability } = 1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right) $$
$$ =1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08 $$
Taking $\mathrm{D}_{(15)}$ as $\frac{15 !}{e}$
$\mathrm{D}_{(14)}$ as $\frac{14 \text { ! }}{e}$
$\mathrm{D}_{(13)}$ as $\frac{13 \text { ! }}{e}$
$$ \text { We get, } \text { Required probability } = 1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right) $$
$$ =1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08 $$
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