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

x 1 + x 2 + x 3 + x 4 = 10 is 80 1296 , x 1 < x 2 and x 3 < x 4 is 25 144 , x 1 < x 2 < x 3 < x 4 is 5 432 , x i ≤ x j for i < j is 7 72

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 ) = 6 C 4 6 4 = 15 1296 B) n(E) = the no.of non-negative integral solutions of the eqn a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 4 ⇒ 9 C 3 = 9 C 4 = 126 C) n(E) = coeft .of x 30 in ⁡ 1 − x 6 ( 1 − x ) − 4 = 9 C 6 − 4 × 1 = 80 D) P ( E ) = 6 C 2 × 6 C 2 6 4 = 25 144

x 1 + x 2 + x 3 + x 4 = 10 is 80 1296

x 1 < x 2 and x 3 < x 4 is 25 144

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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) $P (E) = \frac{6 C_{2} \times 6 C_{2}}{6^{4}} = \frac{25}{144}$