5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is

Question

5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is (A) (1/4) (B) (15/64) (C) (21/64) (D) (17/632)

Tardigrade Admin · Accepted Answer

Total cases =45=1024 Favourable cases =(5 !/(1 !)3 2 !)× (1/3 !)× 4! (By grouping) =240 Hence, required probability =(240/1024)=(15/64)

Q. $5$ different games are to be distributed among $4$ children randomly. The probability that each child gets atleast one game is

$\frac{1}{4}$

$\frac{15}{64}$

$\frac{21}{64}$

$\frac{17}{632}$

Total cases $= 4^{5} = 1024$
Favourable cases $= \frac{5 !}{( 1 ! ) ^{3} 2 !} \times \frac{1}{3 !} \times 4!$ (By grouping)
$= 240$
Hence, required probability
$= \frac{240}{1024} = \frac{15}{64}$

Q. 5 different games are to be distributed among 4 children randomly. The probability that each child gets atleast one game is

41​

6415​

6421​

63217​

Total cases =45=1024 Favourable cases =(1!)32!5!​×3!1​×4! (By grouping) =240 Hence, required probability =1024240​=6415​

Q. $5$ different games are to be distributed among $4$ children randomly. The probability that each child gets atleast one game is

$\frac{1}{4}$

$\frac{15}{64}$

$\frac{21}{64}$

$\frac{17}{632}$

Total cases $= 4^{5} = 1024$
Favourable cases $= \frac{5 !}{( 1 ! ) ^{3} 2 !} \times \frac{1}{3 !} \times 4!$ (By grouping)
$= 240$
Hence, required probability
$= \frac{240}{1024} = \frac{15}{64}$