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Mathematics · Probability

JEE Main 2024 — 8 April, Shift 1 — Question 23

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

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

Blue balls012345
Prob. 5C0⋅ 4C1 9C3\frac{{{~}^{5}}{{\text{C}}_{0}}\cdot {{~}^{4}}{{\text{C}}_{1}}}{{{~}^{9}}{{\text{C}}_{3}}} 5C1 4C2 9C3\frac{{{~}^{5}}{{\text{C}}_{1}}{{~}^{4}}{{\text{C}}_{2}}}{{{~}^{9}}{{\text{C}}_{3}}} 5C2⋅ 4C1 9C3\frac{{{~}^{5}}{{\text{C}}_{2}}\cdot {{~}^{4}}{{\text{C}}_{1}}}{{{~}^{9}}{{\text{C}}_{3}}} 5C3⋅ 4C0 9C3\frac{{{~}^{5}}{{\text{C}}_{3}}\cdot {{~}^{4}}{{\text{C}}_{0}}}{{{~}^{9}}{{\text{C}}_{3}}}00

7x‾=5C14C2+5C2⋅4C1×2+5C3⋅4C0×39C3×77 \overline{\mathrm{x}}=\frac{{ }^{5} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{2} \cdot{ }^{4} \mathrm{C}_{1} \times 2+{ }^{5} \mathrm{C}_{3} \cdot{ }^{4} \mathrm{C}_{0} \times 3}{{ }^{9} \mathrm{C}_{3}} \times 7

30+80+3084×7\frac{30+80+30}{84} \times 7

=14012=706=353=\frac{140}{12}=\frac{70}{6}=\frac{35}{3}

yellow01234
 5C2 4C1{{~}^{5}}{{\text{C}}_{2}}{{~}^{4}}{{\text{C}}_{1}} 5C1 4C2{{~}^{5}}{{\text{C}}_{1}}{{~}^{4}}{{\text{C}}_{2}} 5C0 4C3{{~}^{5}}{{\text{C}}_{0}}{{~}^{4}}{{\text{C}}_{3}}0

4yˉ=40+60+1284×4=11221=1634 \bar{y}=\frac{40+60+12}{84} \times 4=\frac{112}{21}=\frac{16}{3}

7Xˉ+4Yˉ=177 \bar{X}+4 \bar{Y}=17

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Mathematics
Chapter
Probability
Topic
Random Variables, Binomial & Poission Distribution

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