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14.Probability
hard
Two persons each make a single throw with a die. The probability they get equal value is ${p_1}$. Four persons each make a single throw and probability of three being equal is ${p_2}$, then
A
${p_1} = {p_2}$
B
${p_1} < {p_2}$
C
${p_1} > {p_2}$
D
None of these
Solution
(c) ${p_1} = \frac{6}{{36}} = \frac{1}{6}$
To find ${p_2},$ the total number of ways $ = {6^4}$and since two numbers out of $6$ can be selected in ${}^6{C_2}$ ways
$i.e.$ $15$ ways and corresponding to each of these ways, there are $8$ ways e.g., $(1,\,\,1,\,\,1,\,\,2)\,\,(1,\,\,1,\,\,2,\,\,1)\,…..$
Thus favourable ways $ = 15 \times 8 = 120$
Hence ${p_2} = \frac{{120}}{{{6^4}}} = \frac{5}{{54}}$. Hence ${p_1} > {p_2}.$
Standard 11
Mathematics
