Question

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

Answer 1

Answer: (C)

$p _ { 1 } > p _ { 2 }$

Answer 2

$p _ { 1 } = \frac { 6 } { 36 } = \frac { 1 } { 6 }$

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.,

Thus favourable ways $= 15 \times 8 = 120$

Hence $p _ { 2 } = \frac { 120 } { 6 ^ { 4 } } = \frac { 5 } { 54 }$. Hence $p _ { 1 } > p _ { 2 }$