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Probability question

2025 · 2 Apr · Shift 2 · Q45
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Probability question

2025 · 2 Apr · Shift 2 · Q45

JEE MainMathematicsProbabilityMCQ+4 / −1
 Given three indentical bags each containing 10 balls, whose colours are as follows :  Red  Blue  Green  Bag I 325 Bag II 433 Bag III 514\text { Given three indentical bags each containing } 10 \text { balls, whose colours are as follows : } \begin{array}{lccc} & \text { Red } & \text { Blue } & \text { Green } \\ \text { Bag I } & 3 & 2 & 5 \\ \text { Bag II } & 4 & 3 & 3 \\ \text { Bag III } & 5 & 1 & 4 \end{array} Given three indentical bags each containing 10 balls, whose colours are as follows :  Bag I  Bag II  Bag III ​ Red 345​ Blue 231​ Green 534​ A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is qqq, then the value of (1p+1q)\left(\frac{1}{p}+\frac{1}{q}\right)(p1​+q1​) is:
  1. A
    6
  2. B
    9
  3. C
    7
  4. D
    8
View written solutionFree

Correct answer: C

  1. Define events

Let:

  • B1B_1B1​ = chosen bag is Bag I
  • B2B_2B2​ = chosen bag is Bag II
  • B3B_3B3​ = chosen bag is Bag III

Since one bag is chosen at random from 333 bags, P(B1)=P(B2)=P(B3)=13.P(B_1)=P(B_2)=P(B_3)=\frac{1}{3}.P(B1​)=P(B2​)=P(B3​)=31​.

Also let:

  • RRR = drawn ball is red
  • GGG = drawn ball is green
  1. Find p=P(B1∣R)p = P(B_1\mid R)p=P(B1​∣R)

Using Bayes' theorem, P(B1∣R)=P(B1)P(R∣B1)P(R).P(B_1\mid R)=\frac{P(B_1)P(R\mid B_1)}{P(R)}.P(B1​∣R)=P(R)P(B1​)P(R∣B1​)​.

Now, P(R∣B1)=310,P(R∣B2)=410,P(R∣B3)=510.P(R\mid B_1)=\frac{3}{10},\quad P(R\mid B_2)=\frac{4}{10},\quad P(R\mid B_3)=\frac{5}{10}.P(R∣B1​)=103​,P(R∣B2​)=104​,P(R∣B3​)=105​.

So, P(R)=∑i=13P(Bi)P(R∣Bi)P(R)=\sum_{i=1}^3 P(B_i)P(R\mid B_i)P(R)=∑i=13​P(Bi​)P(R∣Bi​) =13(310+410+510)=\frac{1}{3}\left(\frac{3}{10}+\frac{4}{10}+\frac{5}{10}\right)=31​(103​+104​+105​) =13⋅1210=25.=\frac{1}{3}\cdot \frac{12}{10}=\frac{2}{5}.=31​⋅1012​=52​.

Therefore, p=P(B1∣R)=13⋅31025p=P(B_1\mid R)=\frac{\frac{1}{3}\cdot \frac{3}{10}}{\frac{2}{5}}p=P(B1​∣R)=52​31​⋅103​​ =110⋅52=14.=\frac{1}{10}\cdot \frac{5}{2}=\frac{1}{4}.=101​⋅25​=41​.

  1. Find q=P(B3∣G)q = P(B_3\mid G)q=P(B3​∣G)

Again by Bayes' theorem, P(B3∣G)=P(B3)P(G∣B3)P(G).P(B_3\mid G)=\frac{P(B_3)P(G\mid B_3)}{P(G)}.P(B3​∣G)=P(G)P(B3​)P(G∣B3​)​.

Now, P(G∣B1)=510,P(G∣B2)=310,P(G∣B3)=410.P(G\mid B_1)=\frac{5}{10},\quad P(G\mid B_2)=\frac{3}{10},\quad P(G\mid B_3)=\frac{4}{10}.P(G∣B1​)=105​,P(G∣B2​)=103​,P(G∣B3​)=104​.

So, P(G)=13(510+310+410)P(G)=\frac{1}{3}\left(\frac{5}{10}+\frac{3}{10}+\frac{4}{10}\right)P(G)=31​(105​+103​+104​) =13⋅1210=25.=\frac{1}{3}\cdot \frac{12}{10}=\frac{2}{5}.=31​⋅1012​=52​.

Thus, q=P(B3∣G)=13⋅41025q=P(B_3\mid G)=\frac{\frac{1}{3}\cdot \frac{4}{10}}{\frac{2}{5}}q=P(B3​∣G)=52​31​⋅104​​ =215⋅52=13.=\frac{2}{15}\cdot \frac{5}{2}=\frac{1}{3}.=152​⋅25​=31​.

  1. Compute the required value

1p+1q=11/4+11/3=4+3=7.\frac{1}{p}+\frac{1}{q}=\frac{1}{1/4}+\frac{1}{1/3}=4+3=7.p1​+q1​=1/41​+1/31​=4+3=7.

  1. Check options

The value is 777, which corresponds to:

  • Option C

So the correct answer is C.

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