There are three bags B1, B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls, Bags B1, B2 and B3 have probabilities 310,310 and 410 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

(A) Probability that the selected bag is B3 and the chosen ball is green equals 310

(B) Probability that the chosen ball is green equals 3980

(D) Probability that the selected bag is B3, given that the chosen balls is green, equals 513

JEE Advanced · Mathematics · 32. Probability

There are three bags $B_{1}, B_{2}$ and $B_{3} .$ The bag $B_{1}$ contains $5$ red and $5$ green balls, $B_{2}$ contains $3$ red and $5$ green balls, and $B_{3}$ contains $5$ red and $3$ green balls, Bags $B_{1}, B_{2}$ and $B_{3}$ have probabilities $\frac{3}{10}, \frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?

A Probability that the selected bag is $B_{3}$ and the chosen ball is green equals $\frac{3}{10}$
B Probability that the chosen ball is green equals $\frac{39}{80}$
C Probability that the chosen ball is green, given that the selected bag is $B_{3},$ equals $\frac{3}{8}$
D Probability that the selected bag is $B_{3},$ given that the chosen balls is green, equals $\frac{5}{13}$

Verified Solution

Answer & Solution

Correct Answer

(C) Probability that the chosen ball is green, given that the selected bag is $B_{3},$ equals $\frac{3}{8}$

Step-by-step Solution

Detailed explanation

Let us define following events

B_{1}

: Event of selection of bag

1

(

5

Red,

5

Green)

B_{2}

: Event of selection of bag

2

(

3

Red,

5

Green)

B_{3}

: Event of selection of bag

3

(

5

Red,

3

Green)

R =

Event of selection of Red ball

G =

Event of selection of Green ball
Option

(C)

P (\frac{G}{B_{3}}) = \frac{3}{8}

(as bag

3

has been selected now total balls in bag

3 = 8,

green ball

= 3

)
Option

(B)

P (G) = P (B_{1}) P (\frac{G}{B_{1}}) + P (B_{2}) P (\frac{G}{B_{2}}) + P (B_{3}) P (\frac{G}{B_{3}})

= \frac{3}{10} \times \frac{5}{10} + \frac{3}{10} \times \frac{5}{8} + \frac{4}{10} \times \frac{3}{8}

= \frac{39}{80}

Option

(D)

P (\frac{B_{3}}{G}) = \frac{P (B_{3} \cap G)}{P (G)}

= \frac{P (B_{3}) P (\frac{G}{B_{3}})}{P (G)}

= \frac{\frac{4}{10} \times \frac{3}{8}}{\frac{35}{80}} (∴ P (G) = \frac{39}{80})

= \frac{4}{13}

Option

(A)

P (B_{3} \cap G) = P (B_{3}) P (\frac{G}{B_{3}})

P (B_{3} \cap G) = P (B_{3}) P (\frac{G}{B_{3}})

= \frac{3}{20}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbb{R}$, and $g: \mathbb{R} \rightarrow(0, \infty)$ be a function such that $g(x+y)=g(x) g(y)$ for all $x, y \in \mathbb{R}$. If $f\left(\frac{-3}{5}\right)=12$ and $g\left(\frac{-1}{3}\right)=2$, then the value of $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)$ is . ________.

JEE Advanced 2024 Hard

Let

p, q, r

be non-zero real numbers that are, respectively, the

10^{th}, 100^{th}

and

1000^{th}

terms of a harmonic progression. Consider the system of linear equations

x + y + z = 1

10 x + 100 y + 1000 z = 0

q r x + p r y + p q z = 0

	List- $I$		List- $II$
$(I)$	If $\frac{q}{r} = 10$ , then the system of linear equations has	$(P)$	$x = 0, y = \frac{10}{9}, z = - \frac{1}{9}$ as a solution
$(II)$	If $\frac{p}{r} \neq 100$ , then the system of linear equations has	$(Q)$	$x = \frac{10}{9}, y = - \frac{1}{9}, z = 0$ as a solution
$(III)$	If $\frac{p}{q} \neq 10$ , then the system of linear equations has	$(R)$	infinitely many solutions
$(IV)$	If $\frac{p}{q} = 10$ , then the system of linear equations has	$(S)$	no solution
		$(T)$	at least one solution

The correct option is:

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