Probability

Thankyou for the beautiful explanation, Sir. [1]

While applying Baye's theorem one should keep in mind the following. If A_k (k=1,2,..,n) be the set of mutually exclusive and exhaustive events, then the probabilities P(A_k) appearing in the Baye's theorem
P(A_j|A)=\dfrac{P(A_j)P(A|A_j)}{\sum_{k=1}^n P(A_k)P(A|A_k)} for j = 1, 2, ... ,n
are the so called a priori probabilities. That means these are the probabilities of the events A_k without the information supplied by A.

For example, in the given problem, suppose we denote the event
A_k = { the bag contains (k&ndash 1) white balls} for k = 1 to 5
so that we have taken all possible number of white balls in the bag. These five events are equally probable, so that P(A_k) = 1/5.
Suppose the event A = {two white balls are drawn} Then we need P(A₅|A)
The denominator appearing in the Baye's theorem is precisely the probability of A as obtained from the theorem on total probability which just asks to take all possibilities. So finding P(A) first we have

P(A) = P(A₁)P(A|A₁) + P(A₂)P(A|A₂) + P(A₃)P(A|A₃) + P(A₄)P(A|A₄) + P(A₅)P(A|A₅)

Noting that A_k is the event that the bag contains (k&ndash 1) white balls, the above relation gives

P(A)= 0 + 0 + 15 . ²C₂⁴C₂ + 15 . ³C₂⁴C₂ + 15 . ⁴C₂⁴C₂ = 13

So finally applying Baye's formula, we get
P(A₅|A)= P(A₅)P(A|A₅)P(A) = 35
which is the required answer. Note that the final result would remain unchanged even if instead of assume 5 cases A₁ to A₅, we would have assumed only four cases: The bag contains either 1 or 2 or 3 or 4 white balls.

Let A= Two balls drawn are white

E1 = All 4 balls in the box were white
E2 = 3 white,1 non-white
E3= 2 white,2 white
E4 = 3 non white, 1 white
E5 = All non white

P(E1) = 1/16 (Only 1 combination wwww)
P(E2) = 4/16 (4 combinations wwwn,nwww,wwnw,wnww)
P(E3) = 6/16 (6 combinations,similarly)
P(E4) = 4/16 (4 combinations)
P(E5) = 1/16 (1 combinations)

Now, P(A|E1) = 1
P(A|E2) = 3C2 / 4C2 = 1/2
P(A|E3) = 2C2 / 4C2 = 1/6
P(A|E4) = 0
P(A|E5) = 0

By Baye's Theorem,

P(E1|A) = { P(E1) x P(A|E1) } / {P(E1) x P(A|E1) + P(E2) x P(A|E2) +P(E3) x P(A|E3) + P(E4) x P(A|E4) + P(E5) x P(A|E5) }

= { 1x 1/16 } / { 1/16 + 1/8 + 1/16 + 0 + 0}

= 1/4
So, I think 1/4 is the answer.I am still not sure about this but I dont find any mistake in it.
P.S. Acc. to my solution, this ques does not seem to be of boards level.(Provided this is the intended answer)

Now,@shreyan.. how can one assume equal probability of all the five cases???
@rickde.. the probablity of all the cases can't be same.
@sb... yes there are three possible as the other two (P(A|E4) and P(A|E5) are 0
@avik ... u calculated the probability of getting 2 white balls if there are 3w out of 4 as 3/4 * 2/3 .. vich is correct but that is P(A|E3).Should u not multiply it with P(E3) to apply Baye's theorem?

bayes theorem ka q hai

3/5 provided probability of all cases r same

ya avik's answer is correct

Nahi yaar, How can Total Prob. be applied here...? It is Specifically given tht 2Balls have already been drawn & found to be white.

Total ProB. wud have been used if the Qn had been like- "Find the Probability of drawing out 2 White balls frm a bag of 4 unknowns."
.....Tab v wud have taken ProB.(s) of all the possible cases n simply summed up.

@avik liked how confidently u wrote that[1][1]...3/5sahi hai[1]

but just look at ur ncert text or look below::[1]

so total probability being applied here........no chance[1][1]..ha!!!!!!ha!!!!!!!!!

1/4??
since v r not concerned with ne colour but white:
so of remaining 3
1w,2w,3w,no-w
since 3w is favourable 1/4

3/5 Sahi hai.

Possible cases-
i)- All WHite, ii)- 3W, iii)- 2W

Probabilities fr drawing 2W successively frm each,
P(i)=1
P(ii)=3/4*2/3
P(iii)=2/4*1/3

Required= P(i)/[P(i)+P(ii)+P(iii)]=3/5

hey if i go by logic n reason it seems a simple 1/4.[simple use of equiprobability theorem [1][1]]

method-2 [by drawing a sample space for remaining four balls n using class-IX probability formulas..lol!!![3][3][3]]

let W_k denote the event that there are k white balls

A be the event of drawing two white balls from the bag...

we assume P(W_k) = 1/4 for all k 1 - 4

then reqd P(W_4|A)=\frac{1/4}{0+1/4(1/6)+1/4(3/6)+1/4(1)}=\frac{1}{10/6}=\frac{3}{5}

like this perhaps...

how 3/5

3/5 im gettin but sum others were gettin 6/13 ...