Question

I need some quick help with this. I took notes on my class but I don't recall being taught how to do this. Could you please help me? Problem is below: P(A) = ?, P(B) = 0.9, P(A and B) = 0.45

Answer 1

I was thinking I need to set up some type of proportion

Answer 2

is there any more info or background. It seems to be missing some info.

Answer 3

Well, it says that it is an independent event.

Answer 4

Events A and B are independent. Find the indicated probability. Show your work!

a. P(A) = 0.3, P(B) = 0.5, P(A and B) = ?

b. P(A) = ?, P(B) = 0.9, P(A and B) = 0.45

Answer 5

in that case, (for independent events)
P(A & B) = P(A)*P(B)

Answer 6

That's what I did for A, but for B, the value is missing

Answer 7

P(A) = ?, P(B) = 0.9, P(A and B) = 0.45

fill in what you know into
P(A & B) = P(A)*P(B)

Answer 8

For A, I got 3/20, like this: 0.3 = 3/10
0.5 = ½
3/10 x ½ = 3/20

Answer 9

yes for A you should get 0.15 (usually people write props as decimals)

Answer 10

Oh! Okay, I'll be sure to include the decimal too.

Answer 11

.45=.9 * b -->is this right?

Answer 12

For this one
P(A) = ?, P(B) = 0.9, P(A and B) = 0.45

fill in what you know into
P(A & B) = P(A)*P(B)

you get
0.45 = P(A)*0.9

to find P(A), divide both sides by 0.9

Answer 13

you get
\[ \frac{0.45}{0.9}= P(A)\cdot \frac{0.9}{0.9}\]
which simplifies

Answer 14

.45=P(A) * 9
= .5

Answer 15

after dividing both sides by 0.9
\[ \frac{0.45}{0.9}= P(A)\cdot \frac{0.9}{0.9}\]
on the right side 0.9/0.9 is 1 (anything divided by itself is one)
and 1 times P(A) is P(A) (anything times 1 is itself)
so you get
0.5= P(A)
or
P(A)= 0.5 (people usually write it this way)

Answer 16

Okay so what I did is right. I got 0.5 for P(A)

Answer 17

yes, you got it right. But notice how to do it step by step. that is the lesson to try to learn

Answer 18

Thank you so much for your help. I understand. I typed every step on my paper. I'm doing a project, so I'm just going over the ones I truly couldn't understand very well. I only have two more. The rest I did myself. Again, thank you so much for your help.