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Find the probability P(E or F) if E and F are mutually exclusive, P(E)=0.41 and P(F) = 0.47
P(E or F) =

Sagot :

Answer:

P(E or F)=P(E)+P(F)=0.41+0.47=0.88

As per the probability of an event, the probability P(E or F) is 0.88, when E and F are mutually exclusive.

What is probability of an event?

"Probability is simply how likely something is to happen. Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes how likely they are.

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty. "

Given, the probability of the event E is P(E) = 0.41

The probability of the event F is P(F) = 0.47

Now, the probability P(E or F) is

= P(E) + P(F)

= 0.41 + 0.47

= 0.88

Therefore, P(E or F) = 0.88

Learn more about the probability of an event here: https://brainly.com/question/13974679

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