Question

A die is rolled twice and the sum of the numbers appearing is
observed to be 7. What is the conditional probability that the
number 3 has appeared at-least once

Answer 1

The sample space is the set of all combinations of numbers 1-6. (ie S = [(1,1), (1,2), (1,3), ..., (6,5), (6. 6)])

Let E represent the event that the sum of the two die rolls is 7. So E = [(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)] and |E| = 6.

Let T represent the event that three is one of the two rolled numbers. So T = [(3,1), (3,2), ... , (3,6), (1,3), (2,3), ... , (6,3)] and |T| = 12

So the probability of T given E is P(T|E) = 2/6

Answer 2

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If a dice is thrown twice, then the sample space obtained is:

S = {(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

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

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

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

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

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

From the given data, it is needed to find the Probability that 4 has appeared at least once, given the sum of nos. is observed to be 6

Assume that, F: Addition of numbers is 6

and take E: 4 has appeared at least once

So, that, we need to find P(E|F)

Finding P (E):

The probability of getting 4 atleast once is:

E = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)}

Thus , P(E) = 11/ 36

Finding P (F):

The probability to get the addition of numbers is 6 is:

F = {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}

Thus, P(F) = 5/ 36

Also, E ∩ F = {(2,4), (4,2)}

P(E ∩ F) = 2/36

Thus, P(E|F) = (P(E ∩ F) ) / (P (F) )

Now, subsbtitute the probability values obtained= (2/36)/ (5/36)

Hence, Required probability is 2/5

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Hope it's Helpful.....:)