Maximum probablity of both event A & B = HCF of 90 & 20

=> 10

Minimum probablity of both event A & B = LCM of 90 & 20

=> 180

Percentage of A = ( 10 × 100 ) ÷ 100 = 10%

Percentage of B = ( 180 × 100 ) ÷ 100 = 180%

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Verified answer

Answer:

The maximum probability, 0.9, obtains when A happens if B happens. The minimum probability, 0.2, obtains when B happens if A happens.

Step-by-step explanation:

Method 1:

The sum of probabilities of A and B is 1.1, which means the probabilities overlap. Thus events A and B are dependent in some way:

<1> B happens if A happens: P (B | A) = P (A and B) / P(A) = 0.18 / 0.9 = 0.2

<2> A happens if B happens: P (A | B) = P (A and B) / P(B) = 0.18 / 0.2 = 0.9

Therefore, the maximum probability, 0.9, obtains when A happens if B happens. The minimum probability, 0.2, obtains when B happens if A happens.

Method 2:

There are several methods to solve this; however one method is to recognize the mutually exclusive sample space of A∩B, A’∩B, A∩B’, and A’∩B’ must add to 1. If B is wholly contained in A, P(A∩B) = P(A) = 90%; that is the maximum probability. An example is A = any precipitation, B = snowing. If P(A∩B) < 10%, then P(A’∩B) + (A∩B’) > 90%. Therefore P(A∩B) >= 10%, and 10% is the minimum probability. An example is A = sun out, B = raining.

Method 3:

The probability that A occurs is 0.9.

Let the conditional probability that B occurs, given that A occurs, be p.

Then the probability that both A and B occur is 0.9p.

The probability that A does not occur is 0.1.

Let the probability that B occurs, given that A does not occur, be q.

Now p and q are not independent, since the total probability that B occurs is 0.2.

Therefore 0.9p + 0.1q = 0.2

Therefore p = 2/9 - q/9.

The minimum value of p occurs when q = 1, leading to p = 1/9.

The maximum value of p occurs when q = 0, leading to p = 2/9.

Therefore the probability that both A and B occur, being 0.9p, ranges from 0.1 to 0.2.

Reference Link

• https://brainly.in/question/15858063