x: 0, 1, 2, 3; P(x): 0.25, 0.25, 0.25, ?, find the missing probability

Step by step, with the reason for every step · Random variables · University Statistics

Step by step

Question: x can be 0, 1, 2, 3, with probabilities 0.25, 0.25, 0.25, a missing one: find the missing probability.

x: 0, 1, 2, 3; P(x): 0.25, 0.25, 0.25, ?

x always takes one of its values.

So all the probabilities add up to 1.

a probability table always adds up to 1

0.25 plus 0.25 plus 0.25 plus P of 3 is 1.

0.25 + 0.25 + 0.25 + P(3) = 1

The known ones add up to 0.75.

0.75 + P(3) = 1

To remove the plus 0.75, we take away 0.75 on both sides.

1 minus 0.75 is 0.25.

P(3) = 1 − 0.75 = 0.25

Careful!

The missing probability is what is left over from 1.

Careful: the missing P is what is left of 1

Let's check: 0.25 plus 0.25 plus 0.25 plus 0.25 is 1.

So it is right!

0.25 + 0.25 + 0.25 + 0.25 = 1

Let's look back.

First: all P(x) add to 1.

Last: check.

Answer

Careful: the missing P is what is left of 1

Try this one yourself, then check:

X can be 0, 1, 2, 3, with probabilities 0.58, 0.36, 0.01, a missing one: find the x: 0, 1, 2, 3; P(x): 0.58, 0.36, 0.01, ?

Show the answer

P(3) = 0.05