Probablity questions

[robinphilip01]

Hi,
One question:
" suppose a box contains 20 balls . ten ball is white and marked with the integers 1-10. the other 10 balls are red and marked with the integers 11-20. if one ball is selected , what is the probability that the ball will be white or will be marked with an even number ?"

The explnation for the answer 3/8 is like 1/2+1/2-5/20
- 5/20 as the fraction of white and even number marked balls.
My question is do we actully substract the fraction of balls with red and odd number ? ; even though
the fraction is same -5/20 ?

[kucz.adam]

Hi there robinphilip01,

Let's break the solution down into it components:

Step 1: he probability of choosing a white ball (written p(W)) is 10/20 which is 1/2, so we say:

p(W)=1/2

Step 2: The probability of choosing an even numbered ball (p(E)) is 10/20, as the ball can be a 2, 4, 6, 8, 10, 12, 14, 16, 18, or 20.

p(E)=1/2

Step 3:REMEMBER, there are white balls that are evenly numbered, namely ball # 2, 4, 6, 8, and 10. Because of this, there is a particular probability of choosing a ball that is white AND is even numbered p(W n E), which is 5/20, or 1/4.

When you calculate the probability of one event OR another event occurring, you add the probability of each occurring. When these two events are mutually exclusive, you can stop there. When they are not mutually exclusive, you have to account for the possibility that both of these events will occur at the same time by subtracting the probability of this happening.

I hope this has helped!

[tommywallach]

Perfect explanation from Adam. Two other quick things I'd mention. Be careful, because you've written something as negative twice now (on this thread and in another). Don't write -5/20, but 5/20.

And just to be clear, it isn't the odds of "Red and Odd" that we're subtracting, but the odds of "red and even," because we're using the equation:

P(A or B) = P(A) + P(B) - P(A and B)

-t

[robinphilip01]

thankyou..

[tommywallach]

Glad to help!