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Q.Three boxes B1, B2 and B3 contain balls with different colours as shown below:
B1: White 2, Black 1, Red 2
B2: White 3, Black 2, Red 4
B3: White 4, Black 3, Red 2
A die is thrown. B1 is chosen if either 1 or 2 turns up. B2 is chosen if 3 or 4 turns up and B3 is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball drawn is found to be red, find the probability that it is drawn from box B2.

Andhra Pradesh BieapBIEAP Intermediate Board 2026Subjective· 7mImportance★★★★★
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Use Bayes' theorem: P(B2∣Red)=P(B2)P(Red∣B2)∑iP(Bi)P(Red∣Bi)P(B_2\mid\text{Red})=\dfrac{P(B_2)P(\text{Red}\mid B_2)}{\sum_i P(B_i)P(\text{Red}\mid B_i)}.

Box contents and red-ball probabilities:

B1: 2W,1Bk,2R (total 5)⇒P(Red∣B1)=25,B_1:\ 2\text{W},1\text{Bk},2\text{R}\ (\text{total }5)\Rightarrow P(\text{Red}\mid B_1)=\frac25,

B2: 3W,2Bk,4R (total 9)⇒P(Red∣B2)=49,B_2:\ 3\text{W},2\text{Bk},4\text{R}\ (\text{total }9)\Rightarrow P(\text{Red}\mid B_2)=\frac49,

B3: 4W,3Bk,2R (total 9)⇒P(Red∣B3)=29.B_3:\ 4\text{W},3\text{Bk},2\text{R}\ (\text{total }9)\Rightarrow P(\text{Red}\mid B_3)=\frac29.

The die gives each box probability P(B1)=P(B2)=P(B3)=13P(B_1)=P(B_2)=P(B_3)=\dfrac13 (each box corresponds to 2 of the 6 faces).

Total probability of red:

P(Red)=13⋅25+13⋅49+13⋅29=13(25+49+29)=13(25+69)=13(25+23).P(\text{Red}) = \frac13\cdot\frac25+\frac13\cdot\frac49+\frac13\cdot\frac29 = \frac13\left(\frac25+\frac49+\frac29\right) = \frac13\left(\frac25+\frac69\right) = \frac13\left(\frac25+\frac23\right).

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