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Sagot :
The conditional probability, in a single roll of two fair 6 sided dice, that the sum is greater than 6, given that neither die is a two : 17/25
We know that the conditional probability is given by,
P(B | A) = probability of occurrence of event B, given that event A has
occurred
= P(A ∩ B) / P(A)
Here, P(A ∩ B) means the probability of happening two events A and B at the same time.
We also know that if P (B | A ) = P(B) i.e., P(A ∩ B) = P(A) × P(B) the two events A and B are independent of each other.
For this question, let the dice D1 and D2 are rolled once.
Let the numbers displayed on the dice be d1 and d2 respectively.
The dice D1 and D2 are independent.
We need to find the conditional probability that the sum is greater than 6, given that neither die is a two.
Let S represents the sum of the numbers displayed on the dice.
S = d1 + d2
The sum is even, if d1 = d2 is odd OR if d1 = d2 is even
P(d1 = even) = 3/6
=1/2
P(d2=even) = 1/2
P(d1 = odd) = 1/2
P(d2 = odd) = 1/2
So, P(S = even) = [P(d1=even) × P(d2 = even)] + [P(d1= odd) × P(d2=odd)]
= [1/2 × 1/2] + [1/2 × 1/2]
= 1/2
So, we can say that, the sum is either even or odd which are equally likely and hence its probability is 1/2.
First we find the probability for the sum is greater than 6 i.e., P(S > 6)
The possible combination of d1 and d2 for the sum greater than 6 would be,
{(1,6), (2,5), (2, 6), (3, 4), (3, 5), (3, 6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
⇒ n(S > 6) = 21
The number of all possible outcomes = 36
So, P(S > 6) = 21/36
= 7/12
Now we find the probability that neither die is a two
⇒ P(neither die is a two) = [P(1) ∪ P(3 ≤ d1 ≤ 6)] AND [P(1) ∪ P(3 ≤ d1 ≤ 6)]
⇒ P(neither die is a two) = 5/6 × 5/6
⇒ P(neither die is a two) = 25/36
Now, we find the probability that the sum S > 6 AND neither die is a two.
The possible combination for the sum S > 6 AND neither die is a two would be,
{(1,6), (3, 4), (3, 5), (3, 6), (4,3), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,1), (6,3), (6,4), (6,5), (6,6)}
⇒ n(S > 6 AND neither die is a two) = 17
So, P(S > 6 AND neither die is a two) = 17/36
Now we find the conditional probability P(S > 6 | neither die is a two)
⇒ P(S > 6 | neither die is a two) = P(S > 6 AND neither die is a two) ÷
(neither die is a two)
⇒ P(S > 6 | neither die is a two) = (17/36) / (25/36)
⇒ P(S > 6 | neither die is a two) = 17/25
Therefore, the conditional probability, in a single roll of two fair 6 sided dice, that the sum is greater than 6, given that neither die is a two : 17/25
Learn more about the probability here:
https://brainly.com/question/3679442
#SPJ4
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