JEE Advanced · Mathematics · 32. Probability

Suppose that
Box-I contains $8$ red, $3$ blue and $5$ green balls,
Box-ll contains $24$ red, $9$ blue and $15$ green balls,
Box-III contains $1$ blue, $12$ green and $3$ yellow balls,
Box-IV contains $10$ green, $16$ orange and $6$ white balls.
A ball is chosen randomly from Box-l; call this ball $b$ . If $b$ is red then a ball is chosen randomly from Box-ll, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

A $\frac{15}{256}$
B $\frac{3}{16}$
C $\frac{5}{52}$
D $\frac{1}{8}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{5}{52}$

Step-by-step Solution

Detailed explanation

$P (Red from Box I) = \frac{8}{16} = \frac{1}{2}$ $P (Blue from Box I) = \frac{3}{16}$

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

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

If a tangent to a suitable conic (Column 1) is found to be

y = x + 8

and its point of contact is

(8, 16)

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

Which of the following is (are) NOT the square of a $3 \times 3$ matrix with real entries?

JEE Advanced 2017 Easy

Consider the following lists:

	List- $I$		List- $II$
$(I)$	$\{x \in [- \frac{2 π}{3}, \frac{2 π}{3}] : \cos x + \sin x = 1\}$	$(P)$	has two elements
$(II)$	$\{x \in [- \frac{5 π}{18}, \frac{5 π}{18}] : \sqrt{3} \tan 3 x = 1\}$	$(Q)$	has three elements
$(III)$	$\{x \in [- \frac{6 π}{5}, \frac{6 π}{5}] : 2 \cos (2 x) = \sqrt{3}\}$	$(R)$	has four elements
$(IV)$	$\{x \in [- \frac{7 π}{4}, \frac{7 π}{4}] : \sin x - \cos x = 1\}$	$(S)$	has five elements
		$(T)$	has six elements

The correct option is:

JEE Advanced 2022 Medium

Let $S$ be the set of all complex numbers $z$ satisfying $|z^{2} + z + 1| = 1$ . Then which of the following statements is/are TRUE?

JEE Advanced 2020 Hard

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	LIST -I		LIST -II
A)	$d s p^{2}$	P)	${[F e F_{6}]}^{4 -}$
B)	$s p^{3}$	Q)	$[T i {(H_{2} O)}_{3} C l_{3}]$
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D)	$d^{2} s p^{3}$	S)	${[F e C l_{4}]}^{2 -}$
		T)	$N i {(C O)}_{4}$
		U)	${[N i {(C N)}_{4}]}^{2 -}$