Question

A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than $9$, given that the black die resulted in a $5$.
(b) Find the conditional probability of obtaining the sum $8$, given that the red die resulted in a number less than $4$.

• A. $0.33,0.11$
• B. $0.51,0.76$
• C. $0.56,0.43$
• D. $0.11,0.65$

Answer 1

The correct option is A $0.33,0.11$

A black and a red are rolled (a) Find the conditional

Probability of obtaining a sum greater than a, given that

the black die resulted in a 5. (b) Find the conditional

probability of obtaining the sum 8 , given that the red die

resulted in a number less than 4.

Sol (a) Given

There are 2 die's they are block & red

dies are rolled

Then the sample space of both die's

events are $6\times 6=36=S$

Let

$A=\left\{\right(6,6),(6,5),(6,6),(5,5),(4,6),(6,4\left)\right\}$

The farorable outcomes $=6$

$P\left(A\right)=\frac{NooffavourableoutcomesinA}{Totalno.ofoutcomesS}$

$=\frac{6}{36}=\frac{1}{6}$

Let

$B=$ Set of events

Where the black die raled a 5

$B=\left\{\right(1,5),(2,5),(3,5),(4,5),(5,5),(6,5\left)\right\}$

The total favorable outcomes $=6$

$P\left(B\right)=\frac{1}{6}$

outset of events $=\left\{\right(5,5),(6,6\left)\right\}=2$

$P(A\cap B)=\frac{2}{36}$

$\frac{P(A\cap B)}{P\left(B\right)}=\frac{2/36}{1/6}=\frac{2}{6}=\frac{1}{3}$

$=0.333$

(b) Given R = No. on red die is

less than 4.

$S=$ Sum of number is 8

we should find $P\left(R15\right)$

$R=\left\{\right(2,6\left)\right(3,5),(4,4),(15,6\left)\right(6,2\left)\right\}S$

$P\left(R\right)=\frac{5}{36}$

$S=\left\{\right(1,1),(2,1),(3,1),(4,1),(5,1)(6,1),(7,2),(2,2),(3,2)(4,2),(5,2),(6,2),(1,3)(2,3),(3,3),(4,3),(5,3)(6,3)\}$

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

$R\cap S=\left\{\right(5,3),(6,2\left)\right\}$

So $P(R\cap S)=\frac{2}{36}$

$P\left(R/S\right)=\frac{P(R\cap S)}{P\left(S\right)}$

$=\frac{2/36}{18/36}$

$=\frac{2}{18}=\frac{1}{9}$

$=0.1111$