Four persons A,B,C,D roll a die each and the numbers obtained by them are noted as x1 , x2 , x3 , x4 respectively. The probability that

P(E)=6C464=151296B) n(E) = the no.of non-negative integral solutions of the eqna1+a2+a3+a4+a5+a6=4⇒9C3=9C4=126C) n(E) = coeft .of x30in⁡1−x6(1−x)−4=9C6−4×1=80D) P(E)=6C2×6C264=25144

Four persons A,B,C,D roll a die each and the numbers obtained by them are noted as x1 , x2 , x3 , x4 respectively. The probability that

P(E)=6C464=151296B) n(E) = the no.of non-negative integral solutions of the eqna1+a2+a3+a4+a5+a6=4⇒9C3=9C4=126C) n(E) = coeft .of x30in⁡1−x6(1−x)−4=9C6−4×1=80D) P(E)=6C2×6C264=25144

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Four persons A,B,C,D roll a die each and the numbers obtained by them are noted as x₁ , x₂ , x₃ , x₄respectively. The probability that

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$x_{1} + x_{2} + x_{3} + x_{4} = 10 is \frac{80}{1296}$

$x_{1} < x_{2} and x_{3} < x_{4} is \frac{25}{144}$

$x_{1} < x_{2} < x_{3} < x_{4} is \frac{5}{432}$

$x_{i} \leq x_{j} for i < j is \frac{7}{72}$

answer is A, B, C, D.

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Detailed Solution

$P (E) = \frac{6 C_{4}}{6^{4}} = \frac{15}{1296}$
B) n(E) = the no.of non-negative integral solutions of the eqn
$\begin{array}{l} a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 4 \\ \Rightarrow 9_{C_{3}} = 9_{C_{4}} = 126 \end{array}$
C) n(E) = coeft .of $x^{30} in (1 - x^{6}) (1 - x) - 4$
$= 9 C_{6} - 4 \times 1 = 80$
D) $P (E) = \frac{6 C_{2} \times 6 C_{2}}{6^{4}} = \frac{25}{144}$