Question

Two persons each makes a single throw with a die. The probability they get equal value is $p_1.$ Four persons each makes 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. Can't say

Answer 1

Answer: (C)

$p_1>p_2,$

$p_1=\frac{6}{36}=\frac{1}{6}$ [$\because$ Out of total of 36 ways, both the persons can throw equal values in 6 ways ]
To find $p_2,$ the total no. of ways $n=6^4$ and the favourable no. of ways $m=15\times 8=120.$
Since any two numbers our of 6 can be selected in ${}^6{C}_2$ i.e., 15 ways and corresponding to each of these ways, there are 8 ways e.g., corresponding to the numbers 1 and 2 the eight ways are $:$
$(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1),(2,2,2,1),(2,2,1,2),(2,1,2,2),(1,2,2,2).$
Hence $p_2=\frac{120}{6^4}=\frac{5}{54}.$ Since $\frac{1}{6}>\frac{5}{51},$ $p_1>p_2.$