Question

Two dice are thrown together first and secondly three dice are thrown together. The probability that the total in the first throw is 4 or more and at the same time the total in the second throw is 6 or more is $\frac{1130+k}{1296}$. Find the value of $k$.

Answer 1

In the first throw a number less than $4$ can come as

$(1,1),(1,2),(2,1),$ that is in $3$ ways and

So $4$ or more can came in $36$-$3$ i.e.in $33$ ways since two dice can come up in $36$ ways.

Hence the probability of this case $\frac{33}{36}=\frac{11}{12}.$

In the second case, three dice are thrown.

In this case, the total of $3,4$ or $5$ can come as

$(1,1,1).(2,1,1)(1,2,1)(1,1,2),(1,1,3),(3,1,1),(1,3,1),(1,2,2),(2,1,2)and(2,2,1)$

i.e.in $10$ ways.

Hence the number $6$ or more can come in $216$-$10=206$ ways.

Hence the Probability of this case $=\frac{206}{216}=\frac{103}{108}.$

$\therefore$ The required probability $=\frac{11}{12}\times \frac{103}{108}=\frac{1133}{1296}.$