Carcass wrote:

A certain experiment has three possible outcomes. The outcomes are mutually exclusive and have probabilities \(p\), \(\frac{p}{2}\), and \(\frac{p}{4}\), respectively. What is the value of \(p\)?

A) \(\frac{1}{7}\)

B) \(\frac{2}{7}\)

C) \(\frac{3}{7}\)

D) \(\frac{4}{7}\)

E) \(\frac{5}{7}\)

In case anyone is wondering why we need the condition that the outcomes are

mutually exclusive, here's why:

If the outcomes were NOT mutually exclusive, then that would means that some outcomes could occur at the same time, and this would mean that the 3 probabilities need not add to 1.

Here's an example:

Let's say we're rolling 1 standard die (with integers 1 to 6), and we want to examine 3 probabilities:

P(the number is less than 7)

P(the number is less than 8)

P(the number is even)

Notice that ALL 3 of these outcomes can occur at the same time.

The individual probabilities are:

P(the number is less than 7) = 1

P(the number is less than 8) = 1

P(the number is even) = 0.5

As you can see, since the outcomes are NOT mutually exclusive, we can't conclude that the sum of the three probabilities = 1

Cheers,

Brent

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Brent Hanneson – Creator of greenlighttestprep.com

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