• Nov 19th 2009, 07:48 AM
sgk1980
I have the following probability assesment table
no.of blemishes-0,1,2,3,4,5
probability-0.34,0.25,0.19,0.11,0.07,0.04

I was trying to select an event A i.e more than 2 blemishes and event B which is 4 or fewer blemishes.

means P(A>=2) and P(B<=4) inorder to calculate P(A) & P(B)how should I approach?Do I need proceed thru binomial distribution?

• Nov 19th 2009, 01:03 PM
nzmathman
The probability of more than 2 blemishes is just 0.11+0.07+0.04 and probability of 4 or fewer blemishes is 1 - 0.04 = 0.96.
• Nov 19th 2009, 02:41 PM
sgk1980
Thank you for the response. so then A intersection B should be 0.11+0.07 right?
• Nov 22nd 2009, 12:47 AM
nzmathman
Quote:

Originally Posted by sgk1980
Thank you for the response. so then A intersection B should be 0.11+0.07 right?

Yes, since this is the event of 3 or 4 blemishes, just add those probabilities together (Nod)
• Nov 22nd 2009, 08:55 AM
sgk1980
Thanks a lot.One more query if event A has atleast 1 blemish and event B has atmost 3 blemishes then AUB = 0.55(0.25 + 0.11+0.19) right?(Thinking)
• Nov 22nd 2009, 08:58 AM
Plato
Quote:

Originally Posted by sgk1980
I have the following probability assesment table
no.of blemishes-0,1,2,3,4,5
probability-0.34,0.25,0.19,0.11,0.07,0.04

I was trying to select an event A i.e more than 2 blemishes and event B which is 4 or fewer blemishes.

means P(A>=2) and P(B<=4) inorder to calculate P(A) & P(B)how should I approach?Do I need proceed thru binomial distribution?

No indeed.
Just add up the corresponding probabilites from the given table.
• Nov 22nd 2009, 03:33 PM
sgk1980
Thank you .then A complement should be .30 right?