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Box $P$ has $2$ red balls and $3$ blue balls and box $Q$ has $3$ red balls and $1$ blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes $P$ and $Q$ are $\dfrac{1}{3}$ and $\dfrac{2}{3}$ respectively. Given that a ball selected in the above process is a red ball, the probability that it came from the box $P$ is:

1. $\dfrac{4}{19}$

2. $\dfrac{5}{19}$

3. $\dfrac{2}{9}$

4. $\dfrac{19}{30}$

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The probability of selecting a red ball,

$=\left(\dfrac{1}{3}\right)* \left(\dfrac{2}{5}\right)+ \left(\dfrac{2}{3}\right)* \left(\dfrac{3}{4}\right)$

$= \dfrac{2}{15} + \dfrac{1}{2}= \dfrac{19}{30}$

Probability of selecting a red ball from box,

$P=\left(\dfrac{1}{3}\right)* \left(\dfrac{2}{5}\right) = \dfrac{2}{15}$

Given that a ball selected in the above process is
a red ball, the probability that it came from the
box $P$ is $=\left({\dfrac{2}{15} }\div { \dfrac{19}{30}}\right)= \dfrac{4}{19}$
edited

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the formula you wrote is not bayes' theorem..its the forumla for conditional probability
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@satyajeet , i wrote in short form and u can see in next line(last line) write full formula .
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It is Bayes' formula... he sort of even derived it
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I use this method only to solve this type of problem, but pls clear 1 doubt.

for P(A intersection B) we write P(A).P(B)   {in case of independent events}  or P(A).P(B/A) {in case of dependent events}. But how to check which one to use...( i.e. how to determine whether the case is of dependent or independent ) ?

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Normally they given, otherwise we have to infer from data

is P and Q are independent or not ?

if P is rolling a die and Q is tossing a coin, then Find the probability of ( head with die is grater than 2)

in this case P and Q are independent ==> P(P $\cap$ Q) = P(P).P(Q)

Given that a ball selected in the above process is a red ball, the
probability that it came from the box P is:

P(comes from box p / selected ball is red)

= P(comes from box p ∩ selected ball is red) / p (selected ball is red)

here P(comes from box p ∩ selected ball is red) = (1/3)*(2/5) = 2/15

p (selected ball is red) = (1/3)*(2/5) + (2/3) * (3/4) = 19/30

so required probility is = (2/15) / (19/30) = 4/19

Just straight way use bayes theorem you wl get 4/19

Now Applying Baye's Theorem Directly:-

P(P/Red) =[ P(P)* P(Red/P) ] / [ P(P)* P(Red/P)  + P(Q)* P(Red/Q) ]

= ( 1/3 * 2/5 ) / [ (1/3 * 2/5) + (2/3 * 3/4) ]

= 4/19

We can solve this question with the help of probability tables $\rightarrow$

 $Red$ $Blue$ $P$ $\color{red}{\frac{1}{3} \times \frac{2}{5}}$ $\frac{1}{3}\times\frac{3}{5}$ $\frac{1}{3}$ $Q$ $\frac{2}{3} \times \frac{3}{4}$ $\frac{2}{3}\times\frac{1}{4}$ $\frac{2}{3}$ $\color{green}{\frac{19}{30}}$ $\frac{11}{30}$ $1$

Given that a ball selected in the above process is a red ball, the probability that it came from the box $P$ $= \frac{\color{red}{\frac{1}{3} \times \frac{2}{5}}}{\color{green}{\frac{19}{30}}} = \frac{4}{19}$

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