praneethuddagiri wrote:

The events A and B are independent. The probability that both events A and B occur is 0.21. The probability that event A occurs and event B does not occur is 0.49. What is the probability that at least one of the events A and B occur?

a) 0.58

b) 0.79

c) 0.84

d) 0.89

e) 0.93

To organize the data, we can use the following Double-Matrix:

The probability that both events A and B occur is 0.21.

The probability that event A occurs and event B does not occur is 0.49Since (both A and B) = 0.21, and A but not B) = 0.49, we get:

(both A and B) is equal to (A)(B):

(A)(B) = 0.21

(A but not B) is equal to (A)(1-B):

(A)(1-B) = 0.49.

If we divide the first equation by the second, we get:

\(\frac{B}{(1-B)} = \frac{0.21}{0.49}\)

\(\frac{B}{(1-B)} = \frac{21}{49}\)

\(\frac{B}{(1-B)} = \frac{3}{7}\)

\(7B = 3 - 3B\)

\(10B = 3\)

\(B = 0.3\)

If we insert (total B) = 0.3 into the matrix and calculate the remaining values, we get:

Since (neither A nor B) = 0.21, (at least A or B) = 1 - 0.21 = 0.79.

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