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Sagot :
Answer:
1/9 = 0.1111 = 11.11% probability that the first die is a 6 given that the minimum of the two numbers is a 2.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Minimum of the two numbers is 2.
Event B: First die in a 6.
Fair dice:
Each throw has 6 equally likely outcomes. Thus, in total, there are [tex]6^2 = 36[/tex] possible outcomes.
Minimum of the two numbers is a 2.
(2,2), (2,3), (3,2), (2,4), (4,2), (2,5), (5,2), (2,6), (6,2).
9 total outcomes in which the minumum of the two numbers is a two, which means that:
[tex]P(A) = \frac{9}{36}[/tex]
Minimum of the two numbers is a 2, and the first die is a 6.
Only one possible outcome, (6,2). So
[tex]P(A \cap B) = \frac{1}{36}[/tex]
Probability that the first die is a 6 given that the minimum of the two numbers is a 2.
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{36}}{\frac{9}{36}} = \frac{1}{9} 0.1111[/tex]
1/9 = 0.1111 = 11.11% probability that the first die is a 6 given that the minimum of the two numbers is a 2.
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