Four integers, from 0 to 9 inclusive written randomly. Probability that not more than two of them are alike, is

Question

Four integers, from 0 to 9 inclusive written randomly. Probability that not more than two of them are alike, is (A) (504/1000) (B) (956/1000) (C) (963/1000) (D) none

Tardigrade Admin · Accepted Answer

n ( S )=104 n ( A ) =104-( text exactly 3 text alike +1 text different + text all four alike ) = 104-[ 10 C 2 ⋅ 2 ! ⋅ (4 !/3 !)+10] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/84595c756acd9424e379c0584b452ec9-.png" /> p = (1000-37/1000)=(963/1000)

Q. Four integers, from 0 to 9 inclusive written randomly. Probability that not more than two of them are alike, is

$\frac{504}{1000}$

$\frac{956}{1000}$

$\frac{963}{1000}$

none

$n (S) = 1 0^{4}$
$n (A) = 1 0^{4} - (exactly 3 alike + 1 different + all four alike)$
$= 1 0^{4} - [^{10} C_{2} \cdot 2! \cdot \frac{4 !}{3 !} + 10]$

$p = \frac{1000 - 37}{1000} = \frac{963}{1000}$

Q. Four integers, from 0 to 9 inclusive written randomly. Probability that not more than two of them are alike, is

1000504​

1000956​

1000963​

none

n(S)=104 n(A)=104−( exactly 3 alike +1 different + all four alike ) =104−[10C2​⋅2!⋅3!4!​+10] p=10001000−37​=1000963​

$\frac{504}{1000}$

$\frac{956}{1000}$

$\frac{963}{1000}$

$n (S) = 1 0^{4}$
$n (A) = 1 0^{4} - (exactly 3 alike + 1 different + all four alike)$
$= 1 0^{4} - [^{10} C_{2} \cdot 2! \cdot \frac{4 !}{3 !} + 10]$

$p = \frac{1000 - 37}{1000} = \frac{963}{1000}$