A die is thrown twice and the sum of the numbers rising is noted to be 6. Calculate the is the conditional probability that the number 4 has arrived at least once?
Answers
Answer:
the sum of rising the numbers rising is noted to be 6 are (1,5), (2,4), (3,3), (4,2), (5,1)
no. of arriving 4 atleast once are 2
P(E) = 2/5
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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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