Question

Two persons A and B play a game of throwing a pair of dice until one of them wins. A will win if sum of numbers on dice appear to be 6 and B will win. if sum is 7. What is the probability that A wins the game if A starts the game. :

Answer 1

Step-by-step explanation:

A win 7:1,6

6,1

p(A)=

36

6

=

6

1

3,4

4,3

2,5

5,2

B win 10

4,6 P(B)=

36

3

6,4

5,5 =

12

1

B wins ⇒ P(Alosc) P(B win)

+P(Alosc)P(Blose)P(Alosc)P(Bwin)

+...

=

6

5

×

12

1

+

6

5

×

12

11

×

6

5

×

12

1

+...

=

1−

72

55

72

5

=

17

5

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Let a be the First term and r ( < 1) be the Common ratio

Then the sum of the Geometric Progression infinite

number of terms

$\displaystyle \: = \frac{a}{1 - r}$

GIVEN

Two persons A and B play a game of throwing a pair of dice until one of them wins. A will win if sum of numbers on dice appear to be 6 and B will WIN if sum is 7

TO DETERMINE

The probability that A wins the game if A starts the game

EVALUATION

Since two dices are thrown

So the total number of possible outcomes

$= {6}^{2} = 36$

Now A will win if sum of numbers on dice appear to be 6

So

So event points for A is

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

So the total number of possible outcomes for A is 5

Again B will win if sum is 7

So the event points for B is

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

So the total number of possible outcomes for B is 6

The X be the event that A wins the game if A starts the game

So

$X = A \cup( A' \cap \: B' \cap \: A ) \cup \: ( A' \cap \: B' \cap \: A' \cap \: B' \cap \: A \: ) + ........$

So the required probability is

$P(X) =P( A ) +P ( A' \cap \: B' \cap \: A ) +P ( A' \cap \: B' \cap \: A' \cap \: B' \cap \: A \: ) + ........$

$\implies \: \displaystyle \: P(X) = \frac{5}{36} +( \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36} ) + ( \frac{31}{36} \times \frac{5}{6} \times \frac{31}{36} \times \frac{5}{6} \times \frac{5}{36} )) + ....$

$\implies \: \displaystyle \: P(X) = \frac{5}{36} \bigg( \: ( \frac{31}{36} \times \frac{5}{6}) + { ( \frac{31}{36} \times \frac{5}{6} ) }^{2} \: + ...... \bigg)$

$\implies \: \displaystyle \: P(X) = \frac{5}{36} \bigg( \frac{1}{1 -\frac{31}{36} \times \frac{5}{6} } \bigg)$

$\implies \: \displaystyle \: P(X) = \frac{5}{36} \times \bigg( \frac{216}{216 - 155} \bigg)$

$\implies \: \displaystyle \: P(X) = \frac{5}{36} \times \bigg( \frac{216}{61} \bigg)$

$\implies \: \displaystyle \: P(X) = \frac{30}{61}$

RESULT

SO THE REQUIRED PROBABILITY IS

$\displaystyle \: \frac{30}{61}$