JEE Advanced · Mathematics · 32. Probability

Let $n_{1} a n d n_{2}$ be the number of red and black balls, respectively, in box I. Let $n_{3} a n d n_{4}$ be the number of red and black balls, respectively, in box II.
One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The balls was found to be red. If the probability that this red ball was drawn from box II is $\frac{1}{3}$ , then the correct option(s) with the possible values of $n_{1}, n_{2}, n_{3} a n d n_{4}$ is(are)

A $n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15$
B $n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50$
C $n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20$
D $n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20$

Verified Solution

Answer & Solution

Correct Answer

(B) $n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50$

Step-by-step Solution

Detailed explanation

\frac{1}{3} = \frac{\frac{1}{2} (\frac{n_{3}}{n_{3} + n_{4}})}{\frac{1}{2} (\frac{n_{1}}{n_{1} + n_{2}} + \frac{n_{3}}{n_{3} + n_{4}})}

\frac{1}{3} = \frac{(\frac{n_{3}}{n_{3} + n_{4}})}{(\frac{n_{1}}{n_{1} + n_{2}} + \frac{n_{3}}{n_{3} + n_{4}})}

(i)

R H S \frac{\frac{5}{20}}{\frac{5}{20} + \frac{3}{6}} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1 + 2}{4}} = \frac{1}{3}

(ii)

R H S \frac{\frac{10}{60}}{\frac{10}{60} + \frac{3}{9}} = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{3}} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}

(iii)

\frac{\frac{5}{25}}{\frac{8}{14} + \frac{5}{25}} = \frac{\frac{1}{5}}{\frac{4}{7} + \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{20 + 7}{35}} = \frac{7}{27}

(iv)

\frac{\frac{5}{25}}{\frac{6}{18} + \frac{5}{25}} = \frac{\frac{1}{5}}{\frac{5 + 3}{15}} = \frac{\frac{1}{5}}{\frac{8}{15}} = \frac{1}{5} \times \frac{15}{8} = \frac{3}{8}

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

Column I		Column II
$(A)$	In a triangle $∆ X Y Z$ let $a, b$ and $c$ be the lengths of the angles $X, Y$ and $Z$ , respectively. If $2 (a^{2} - b^{2}) = c^{2}$ and $λ = \frac{\sin (X - Y)}{sinZ}$ then possible values of $n$ for which $\cos (n π λ) = 0$ is are	$(P)$	$1$
$(B)$	In a triangle $Δ X Y Z,$ let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$ , respectively. If $1 + \cos 2 x - 2 \cos 2 y$ $= 2 \sin x \sin y$ , then possible value(s) of $\frac{a}{b}$ is (are)	$(Q)$	$2$
$(C)$	In $R^{2},$ let $\sqrt{3} \hat{i} + \hat{j}, \hat{i} + \sqrt{3} \hat{j}$ and $β \hat{i} + (1 - β) \hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$ , respectively. If the distance of $Z$ from the bisector of the acute angle of $\vec{O X}$ with $\vec{O Y}$ is $\frac{3}{\sqrt{2}}$ , then possible value (s) of $\|β\|$ is (are)	$(R)$	$3$
$(D)$	Suppose that $F (α)$ denotes the area of the region bounded by $x = 0, x = 2, y^{2} = 4 x$ and $y = \|a x - 1\| +$ $\|α x - 2\| + α x,$ where $α \in \{0, 1\} .$ Then the value (s) of $F (α) + \frac{8}{3} \sqrt{2},$ when $α = 0$ and $α = 1$ , is (are)	$(S)$	$5$
		$(T)$	$6$

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Match List I with List II and select the correct answer using the code given below the lists :

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A.	${(\frac{1}{y^{2}} (\frac{cos ({tan}^{- 1} y) + y sin ({tan}^{- 1} y)}{cot ({sin}^{- 1} y) + tan ({sin}^{- 1} y)}) + y^{4})}^{1 / 2} takes value$	P.	$\frac{1}{2} \sqrt{\frac{5}{3}}$
B.	If cosx + cosy + cosz = 0 = sinx + siny + sinz then possible value of cos $\frac{x - y}{2} is$	Q.	$\sqrt{2}$
C.	If cos $(\frac{π}{4} - x)$ cos2x + sinx sin 2x secx = cosx sin2x secx + cos $(\frac{π}{4} + x)$ cos2x then possible value of secx is	R.	$\frac{1}{2}$
D.	If cot $({sin}^{- 1} \sqrt{1 - x^{2}}) = sin ({tan}^{- 1} (x \sqrt{6})), x \neq 0,$ then possible value of x is	S.	1

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