Verified Solution • 2025–26 Session moderate

QJEEM24Maf086: JEE Main 2024 Mathematics - Class 12 JEEM24Maf086 Solution: Three balls are drawn at random from a bag containing 5 blue…

Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables XX and YY respectively denote the number of blue and yellow balls. If Xˉ\bar{X} and Yˉ\bar{Y} are the means of XX and YY respectively, then 7Xˉ+4Yˉ7 \bar{X}+4 \bar{Y} is equal to ___________.

Step-by-Step Verified Solution

XX 3 2 1 0
YY 0 1 2 3

amp;Xˉ=∑Xp(X)amp;Yˉ=∑Yp(Y)amp;P(X=3)=P(Y=0)=5C3⋅C09C3=5C29C3=542amp;P(X=2)=P(Y=1)=5C2⋅C19C3=1021amp;P(X=1)=P(Y=2)=5C1⋅C29C3=514amp;P(X=0)=P(Y=3)=5C0⋅C39C3=484=121amp;Xˉ=3×542+2×1021+514+0×121=15+40+1542=7042amp;Yˉ=0×542+1×1021+2×514+3×121=20+30+642=5642amp;7Xˉ+4Yˉ=17\begin{aligned}& \bar{X}=\sum X p(X) \\& \bar{Y}=\sum Y p(Y) \\& P(X=3)=P(Y=0)=\frac{{ }^5 C_3 \cdot C_0}{{ }^9 C_3}=\frac{{ }^5 C_2}{{ }^9 C_3}=\frac{5}{42} \\& P(X=2)=P(Y=1)=\frac{{ }^5 C_2 \cdot C_1}{{ }^9 C_3}=\frac{10}{21} \\& P(X=1)=P(Y=2)=\frac{{ }^5 C_1 \cdot C_2}{{ }^9 C_3}=\frac{5}{14} \\& P(X=0)=P(Y=3)=\frac{{ }^5 C_0 \cdot C_3}{{ }^9 C_3}=\frac{4}{84}=\frac{1}{21} \\& \bar{X}=3 \times \frac{5}{42}+2 \times \frac{10}{21}+\frac{5}{14}+0 \times \frac{1}{21}=\frac{15+40+15}{42}=\frac{70}{42} \\& \bar{Y}=0 \times \frac{5}{42}+1 \times \frac{10}{21}+2 \times \frac{5}{14}+3 \times \frac{1}{21}=\frac{20+30+6}{42}=\frac{56}{42} \\& 7 \bar{X}+4 \bar{Y}=17\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring