A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is

Question

A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is (A) (48/91) (B) (44/91) (C) (88/91) (D) (24/91)

Tardigrade Admin · Accepted Answer

P(E) = P(R R B W or B B R W or W W R B) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a7681dd70263c84917901f16e596d9af-.png" /> n ( E )= 6 C 2 ⋅ 5 C 1 ⋅ 4 C 1+ 5 C 2 ⋅ 6 C 1 ⋅ 4 C 1+ 4 C 2 ⋅ 6 C 1 ⋅ 5 C 1 n ( S )= 15 C 4 ∴ P ( E )=(720 ⋅ 4 !/15 ⋅ 14 ⋅ 13 ⋅ 12)=(48/91)

Q. A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is

$\frac{48}{91}$

$\frac{44}{91}$

$\frac{88}{91}$

$\frac{24}{91}$

$P (E) = P (RRB W$ or $BBR W$ or $WW RB)$

$n (E) =^{6} C_{2} \cdot^{5} C_{1} \cdot^{4} C_{1} +^{5} C_{2} \cdot^{6} C_{1} \cdot^{4} C_{1} +^{4} C_{2} \cdot^{6} C_{1} \cdot^{5} C_{1}$
$n (S) =^{15} C_{4}$
$∴ P (E) = \frac{720 \cdot 4 !}{15 \cdot 14 \cdot 13 \cdot 12} = \frac{48}{91}$

Q. A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is

9148​

9144​

9188​

9124​

P(E)=P(RRBW or BBRW or WWRB) n(E)=6C2​⋅5C1​⋅4C1​+5C2​⋅6C1​⋅4C1​+4C2​⋅6C1​⋅5C1​ n(S)=15C4​ ∴P(E)=15⋅14⋅13⋅12720⋅4!​=9148​

$\frac{48}{91}$

$\frac{44}{91}$

$\frac{88}{91}$

$\frac{24}{91}$

$P (E) = P (RRB W$ or $BBR W$ or $WW RB)$

$n (E) =^{6} C_{2} \cdot^{5} C_{1} \cdot^{4} C_{1} +^{5} C_{2} \cdot^{6} C_{1} \cdot^{4} C_{1} +^{4} C_{2} \cdot^{6} C_{1} \cdot^{5} C_{1}$
$n (S) =^{15} C_{4}$
$∴ P (E) = \frac{720 \cdot 4 !}{15 \cdot 14 \cdot 13 \cdot 12} = \frac{48}{91}$