Fifteen football players of a elub-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

Question

Fifteen football players of a elub-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 (A) (5/36) (B) (2/15) (C) (5/24) (D) (1/6)

Tardigrade Admin · Accepted Answer

Required probability =1-(D(15)+ 15 C1 ⋅ D(14)+ 15 C2 D(13)/15 !) Taking D (15) as (15 text ! /e) D (14) as (14 text ! /e) D (13) as (13 !/e) We get, 1-(((15 !/e)+15 ⋅ (14 !/e)+(15 × 14/2) × (13 !/e)/15 !)) =1-((1/e)+(1/e)+(1/2 e))=1-(5/2 e) ≈ .08

Q. Fifteen football players of a elub-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

$\frac{5}{36}$

$\frac{2}{15}$

$\frac{5}{24}$

$\frac{1}{6}$

Q. Fifteen football players of a elub-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

365​

152​

245​

61​

Required probability =1−15!D(15)​+15C1​⋅D(14)​+15C2​D(13)​​ Taking D(15)​ as e15 ! ​ D(14)​ as e14 ! ​ D(13)​ as e13!​ We get, 1−(15!e15!​+15⋅e14!​+215×14​×e13!​​) =1−(e1​+e1​+2e1​)=1−2e5​≈.08

$\frac{5}{36}$

$\frac{2}{15}$

$\frac{5}{24}$

$\frac{1}{6}$