# A student takes examination in four subjects A, B, C, D. His chance of passing in A is 4/5,in B is 3/4in C is 5/6 and in D is 2/3.To qualify he must pass in A and at least two other subjects. What is the probability that he qualifies?

## To find the probability that the student qualifies, we need to calculate the probability of all possible cases where he will pass in A and at least two other subjects.

Case 1: Passing in A, B, and C (but failing in D)
The probability of passing in A, B, and C and failing in D can be calculated as:
P(A) * P(B) * P(C) * (1 - P(D))
= (4/5) * (3/4) * (5/6) * (1 - 2/3)
= (4/5) * (3/4) * (5/6) * (1/3)
= 1/5

Case 2: Passing in A, C, and D (but failing in B)
The probability of passing in A, C, and D and failing in B can be calculated as:
P(A) * P(C) * P(D) * (1 - P(B))
= (4/5) * (5/6) * (2/3) * (1 - 3/4)
= (4/5) * (5/6) * (2/3) * (1/4)
= 1/5

Case 3: Passing in A, B, and D (but failing in C)
The probability of passing in A, B, and D and failing in C can be calculated as:
P(A) * P(B) * P(D) * (1 - P(C))
= (4/5) * (3/4) * (2/3) * (1 - 5/6)
= (4/5) * (3/4) * (2/3) * (1/6)
= 1/10

Adding up the probabilities of these three cases gives us the probability that the student qualifies:
P(qualifies) = P(A,B,C) + P(A,C,D) + P(A,B,D)
= 1/5 + 1/5 + 1/10
= 2/10 + 2/10 + 1/10
= 5/10

Therefore, the probability that the student qualifies is 5/10 or 1/2.

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## To find the probability that the student qualifies, we need to consider different scenarios.

First, let's calculate the probability of passing in exactly two subjects (excluding subject A). We can pass in subjects B and C, subjects B and D, and subjects C and D.

Probability of passing in B and C:
P(B) = 3/4
P(C) = 5/6
P(not B) = 1 - P(B) = 1 - 3/4 = 1/4
P(not C) = 1 - P(C) = 1 - 5/6 = 1/6

The probability of passing in B and C is: P(B) * P(C) * P(not A) * P(not D) = (3/4) * (5/6) * (1/4) * (1/6)

Similarly, we can calculate the probabilities of passing in B and D, and C and D:

Probability of passing in B and D:
P(D) = 2/3
P(not D) = 1 - P(D) = 1 - 2/3 = 1/3

The probability of passing in B and D is: P(B) * P(D) * P(not A) * P(not C) = (3/4) * (2/3) * (1/4) * (1/6)

Probability of passing in C and D:
P(C) = 5/6
P(D) = 2/3
P(not C) = 1 - P(C) = 1 - 5/6 = 1/6

The probability of passing in C and D is: P(C) * P(D) * P(not A) * P(not B) = (5/6) * (2/3) * (1/4) * (1/3)

Now, let's calculate the probability of passing in all three subjects (B, C, and D):
P(B) = 3/4
P(C) = 5/6
P(D) = 2/3
P(not B) = 1 - P(B) = 1 - 3/4 = 1/4
P(not C) = 1 - P(C) = 1 - 5/6 = 1/6
P(not D) = 1 - P(D) = 1 - 2/3 = 1/3

The probability of passing in all three subjects is: P(B) * P(C) * P(D) * P(not A) = (3/4) * (5/6) * (2/3) * (1/4)

Finally, let's calculate the probability of passing in all four subjects (including subject A):
P(A) = 4/5
P(B) = 3/4
P(C) = 5/6
P(D) = 2/3

The probability of passing in all four subjects is: P(A) * P(B) * P(C) * P(D) = (4/5) * (3/4) * (5/6) * (2/3)

To find the probability that the student qualifies, we need to add up the probabilities of all the scenarios where he passes in at least two subjects:

Probability of qualifying = (P(B) * P(C) * P(not A) * P(not D)) + (P(B) * P(D) * P(not A) * P(not C)) + (P(C) * P(D) * P(not A) * P(not B)) + (P(B) * P(C) * P(D) * P(not A)) + (P(A) * P(B) * P(C) * P(D))

Probability of qualifying = [(3/4) * (5/6) * (1/4) * (1/6)] + [(3/4) * (2/3) * (1/4) * (1/6)] + [(5/6) * (2/3) * (1/4) * (1/3)] + [(3/4) * (5/6) * (2/3) * (1/4)] + [(4/5) * (3/4) * (5/6) * (2/3)]

Now, add up all these probabilities and simplify to find the final answer.

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## p(not pass B, C, D) = 1/4 * 1/6 * 1/3 = 1/72

p(not pass B, C) = 1/4 * 1/6 = 1/24 = 3/72

p(not pass B, D) = 1/4 * 1/3 = 1/12 = 6/72

p(not pass C, D) = 1/6 * 1/3 = 1/18 = 4/72

p(pass at least two) = 1 - 14/72 = 58/72

p(qualify) = 4/5 * 58/72