Bayes' Theorem? 'Let S be a region divided into 4 pairs...'

Let S be a region divided into four pairs - 50, 30, 12 and 8 %. A spot falls on the region S, one of the four parts, and an event A occurs with probability 0.01, 0.05, 0.2 and 0.5. If the event A becomes which pair has the biggest probability of having the spot.

Re: Bayes' Theorem? 'Let S be a region divided into 4 pairs.

N1CKY said:

Let S be a region divided into four pairs - 50, 30, 12 and 8 %. A spot falls on the region S, one of the four parts, and an event A occurs with probability 0.01, 0.05, 0.2 and 0.5. If the event A becomes which pair has the biggest probability of having the spot.

Click to expand...

What is meant by dividing a region (an area) in "pairs"?

Are all the four listed numbers percentages (so "percent" signs should follow each one, not just the last)?

Are the four listed numbers just two of the "pairs"? If so, what are the other two pairs?

How does A relate to S? Are the probabilities those of landing on each "part"? :?:

Note: A portion of text appears to have been omitted. "If the event A becomes..." Becomes what? "...of having the spot?" What "spot"?

Please reply with clarification. When you reply, please show all of your work and reasoning so far. Thank you!

Eliz.

If "pair" = "part" and "becomes" = "occurs", then look for the largest joint probability of having the spot, i.e., max of Prob(spot on part i)Prob(A|spot on part i). Note that the max posterior probabilities must be in the same order, so you need not even calculate the marginal probability of A.

Ok let pair be PART

The event A becomes TRUE!

occurs = become.

Here is what i have found so far:

Let S_j be the regions ( j=1,2,3,4 )
Let A_j is the even that the spot falls on Sj

P(A_1)=1/2,
P(A_2)=3/10,
P(A_3)=3/25,
P(A_4)=2/25.

P(A|A_1)=1/100,
P(A|A_2)=1/20,
P(A|A_3)=1/5,
P(A|A_4)=1/2

k_j:=P(A_j|A).P(A)=P(A|A_j).P(A_j)


Seem like there is no point of estimating P(A)



Is that true? Or i need to make corrections?