Question

A fair die is rolled four times. If the probability that outcome of each of the roll is at least as large as the outcome of the preceding roll, can be expressed as $\frac{a}{bc}$, where $a$ is natural number and $bc$ is two digit number, then

• A. $a+b+c=15$
• B. $c+a÷b=3$
• C. $a=b$
• D. $3c-a÷b=4$

Answer 1

Answer: (B), (C)

$a=b$

Let $N_i$ represent the number appearing on die when die is rolled $i^{\text{th}}$ time, $i=1,2,3,4$
Given that
$N_1\le N_2\le N_3\le N_4$
Each time when a die is rolled, six options are there.
Let we have four equal sets $\{1,2,3,4,5,6\}$, now we have to select four numbers such that, exactly one number is selected from each set

Case $1:$ All four are same
Number of ways: ${}^6{C}_1=6$

Case $2:$ Three are same but one is different
Number of ways: ${}^6{C}_1{\times }^5{C}_1=30$

Case $3:$ Two are same of one kind and two are same of second kind
Number of ways: $\frac{1}{2}\times {}^6{C}_1{\times }^5{C}_1=15$

Case $4:$ Two are same and two are different
Number of ways: ${}^6{C}_1{\times }^5{C}_2=60$

Case $5:$ All four are different
Number of ways: ${}^6{C}_4=15$

Required probability $=\frac{6+30+15+60+15}{6\times 6\times 6\times 6}$
$=\frac{126}{36\times 36}=\frac{7}{72}=\frac{a}{bc}$
$a=7,b=7,c=2$
$\therefore a=b$
and $c+a÷b=3$
$2+\frac{7}{7}=3$