Question

In successive rolls of a pair of fair dice, what is the probability of getting $2$sevens before $6$even numbers?

Short answer

The probability of getting 2 sevens before 6 even numbers are$55.5\%$

Step-by-step solution

Given Information

what is the probability of getting $2$ sevens before even numbers

Step 2: Explanation

The probability that two sevens will be rolled before six-event numbers in a sequence of rolls of two dice

If any number is rolled that is not $7$or even, it will be dismissed because it doesn't effect the probability in question

Name

C - $7$or even number is rolled

S - a $7$is rolled

E - an even number is rolled

Step 3: Explanation

P(C)

There are $36$equally likely results of a roll of two dice: Twelve of which do not sum up to $7$or even number-

$(1,2),(2,1),(1,4),(2,3),(3,2),(4,1),(3,6),(4,5),(5,4),(6,3),(5,6),(6,5)$

The remaining $24$form events c

$P\left(C\right)=\frac{24}{36}=\frac{2}{3}$

P(S)

Six of those $36$equally likely events are that the sum is $7$

$P\left(S\right)=\frac{6}{36}$

P(E)

Since $C=E\cup S$

$P\left(E\right)=\frac{18}{36}$

Now the definition of conditional independence yields.

$P_C\left(S\right)=P(S\mid C)=\frac{P\left(SC\right)}{P\left(C\right)}\stackrel{S\subseteq C}{=}\frac{P\left(S\right)}{P\left(C\right)}=\frac{6}{24}=\frac{1}{4}$

$P_C\left(E\right)=P(E\mid C)=\frac{P\left(EC\right)}{P\left(C\right)}\stackrel{E\subseteq C}{=}\frac{P\left(E\right)}{P\left(C\right)}=\frac{18}{24}=\frac{3}{4}$

Explanation

$\text{After seven}C\text{rolls, either}6\text{even numbers or}2\text{sevens have been rolled. And precisely one of those happened.}$

The probability that two sevens were rolled before six even numbers is the probability that they have been rolled in the first $7$important rolls.

Compute first the probability of the complement.

$\begin{array}{r}{P}_C("\text{six or seven even numbers are rolled"})=P_C("6\text{even numbers are rolled"})+\\ P_C("7\text{even numbers are rolled" )}\end{array}$

$=7{\left[P_C\right(E\left)\right]}^6{P}_C\left(S\right)+{\left[P_C\right(E\left)\right]}^7$

$={\left(\frac{3}{4}\right)}^6[\frac{7\cdot 1}{4}+\frac{3}{4}]$

$=\frac{3^6\cdot 10}{4^7}$

$P_C("\text{at least two sevens are rolled")}=1-P_C("\text{six or seven even numbers are rolled")}$

$=1-\frac{3^6\cdot 10}{4^7}$

$=55.5\%$

Final Answer

The probability of getting $2$sevens before $6$even numbers are$=55.5\%$