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

Hint : Conditional probability i.e. $P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$, this formula can also be used when we have to find $P(A \cup B)$. Here two dice are rolled at a time so the total number of outcomes is 36 by keeping these two points in mind we can solve this question very easily.

Complete step by step solution :
There are 2 dice black and red rolled at a time.
So, the total sample space of both the dices are $6 \times 6 = 36$
Now first of all solve the first part of the question.
( i ) It is said that the sum of both the dies must be greater than 9 . So, let us find all the possible outcomes on both dies whose sum is greater than 9 and name this probability as
P ( A ).
So, now A = \left\\{ {(6,6),(6,5),(6,4),(5,6),(5,5),(4,6)} \right\\}
So, by this we get 6 favourable outcomes
P(A) = $\dfrac{{Number\,of\;favourable\;outcome}}{{total\;sample\;space}}$
P(A) = $\dfrac{6}{{36}}\; = \;\dfrac{1}{6}$
Now it is given that the black dice always resulted in 5. So, now let us find the possible outcome of getting 5. And let these outcomes be named B.
So, now B = \left\\{ {5,1),(5,2),(5,3),(5,4),(5,5),(5,6)} \right\\}
Again total favourable outcome = 6
So, now P (B) = $\dfrac{6}{{36}}\; = \;\dfrac{1}{6}$
And it is also said that the sum of both dice must be greater than 9 and the black die must also result in 5.
So, now let us find the common outcomes which have 5 on black dice and whose sum is greater than 9.
And such common events are$A \cap B$= \left\\{ {(5,5),(6,5)} \right\\}= 2
$P(A \cap B) = \dfrac{{Total\;common\;probability}}{{total\;sample\;space}}$ $= \dfrac{2}{{36}} = \dfrac{1}{{18}}$
Now as per given condition there must be 5 on black die. It means P(B) is fixed.
According to conditional probability $P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$
$P(A|B) = \dfrac{{1/18}}{{1/6}} = \dfrac{6}{{18}} = \dfrac{1}{3}$ = 0.33
So, by this we get the conditional probability for part ( i ) as 0.33.
Now similarly solving part (ii)
(ii) It is said that the sum of both the dies must be 8. So, let us find all the possible outcomes on both dies whose sum is 8 and name this probability as P (X)
So, now X = \left\\{ {(2,6),(3,5),(4,4),(5,3),(6,2)} \right\\}
So, by this we get 6 favourable outcomes
P(X) = $\dfrac{{Number\,of\;favourable\;outcome}}{{total\;sample\;space}}$ = $\dfrac{5}{{36}}\;$
Now it is given that the red dice always resulted in a number less than 4. So, now let us find the possible outcome of getting a number less than 4. And let these outcomes be named Y.
So, now Y = \left\\{ {(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3)} \right\\}
Again total favourable outcome = 18
So, now P(Y) = $\dfrac{{18}}{{36}}\; = \;\dfrac{1}{2}$
And it is also said that the sum of both dice must be 8 and the red die must also result in a number less than 4.
So, now let us find the common outcomes which have number less than 4 on red dice and whose sum is 8.
And such common events are$X \cap Y$= \left\\{ {(5,3),(6,2)} \right\\}= 2
P(X \cap Y) = \dfrac{{Total\;common\;probability}}{{total\;sample\;space}}$$$$ = \dfrac{2}{{36}} = \dfrac{1}{{18}}
Now as per given condition there must be a number less than 4 on red die. It means P(Y) is fixed.
According to conditional probability $P(X|Y) = \dfrac{{P(X \cap Y)}}{{P(Y)}}$
$P(X|Y) = \dfrac{{2/36}}{{1/2}} = \dfrac{2}{{18}} = \dfrac{1}{9}$ = 0.11
So, by this we get the conditional probability for part ( ii ) as 0.11.
Hence A is the correct option.

Note : Whenever we come up with this type of problem than we must know that in the formula of conditional probability the common or we can say that the intersection point must kept on the numerator whereas the denominator must be that part which is said to be compulsory, or fixed and these two things are the two most important part of this type of questions.