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If a and b are equally likely events and we require that the probability of their intersection be at least .98, what is p(

a.?

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Final answer:

To find the probability of events a and b, we can use the equation P(a AND b) = P(a) * P(b), and solve for P(a).

Step-by-step explanation:

In this question, we are given that events a and b are equally likely, and we are required to find the probability of their intersection.

Since a and b are equally likely, we can assume that P(a) = P(b).

Let x be the probability of event a. Then, the probability of event b is also x.

Since the probability of their intersection is at least 0.98, we have:

P(a AND b) = P(a) * P(b) >= 0.98

Since P(a) = P(b) = x, we substitute to get:

x * x >= 0.98

x^2 >= 0.98

Taking the square root of both sides:

x >= sqrt(0.98)

Therefore, the probability of event a (P(a)) is greater than or equal to the square root of 0.98.

User Mitya Ustinov
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4 votes

If A and B are equally likely events such that the probability of their intersection be at least 0.98, than we can find P(A) as follows;

We know by DeMorgan's law we have,


\\ P(A\cap B)=1-P(\overline{A\cap B})\\ = 1-P(\overline{A}\cap \overline{B})\\ \text{Since,} P(\overline{A}\cap \overline{B})\leq P(\overline{A})+P(\overline{B}), P(A\cap B)\geq 1-P(\overline{A})-P(\overline{B})\\ \text{we know that,}P(\overline{A})=P(\overline{B})\text{, than we have;}\\ P(A\cap B)\geq 1-2P(\overline{A})\geq 0.98\\ \Rightarrow P(A)\geq 0.99

User Hthms
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