JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Probability

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In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability

\frac{3}{4}

and the remaining 6 questions correctly with probability

\frac{1}{4}

.

If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is

\frac{27 k}{4 ^{10}}

, then k is equal to ____.

Answer: 479

Solution

Let set A = {questions 1,2,3,4} with P(correct) = 3/4, and set B = {questions 5,6,7,8,9,10} with P(correct) = 1/4.

For exactly 8 correct answers, we need:

(4 from A, 4 from B): $(4 4) (\frac{3}{4})^{4} \cdot (4 6) (\frac{1}{4})^{4} (\frac{3}{4})^{2}$
(3 from A, 5 from B): $(3 4) (\frac{3}{4})^{3} (\frac{1}{4}) \cdot (5 6) (\frac{1}{4})^{5} (\frac{3}{4})$
(2 from A, 6 from B): $(2 4) (\frac{3}{4})^{2} (\frac{1}{4})^{2} \cdot (6 6) (\frac{1}{4})^{6}$

Total probability:

P = \frac{1}{4 ^{10}} [3^{4} \cdot 15 \cdot 3^{2} + 4 \cdot 3^{3} \cdot 6 \cdot 3 + 6 \cdot 3^{2}]

= \frac{1}{4 ^{10}} [81 \cdot 15 \cdot 9 + 4 \cdot 27 \cdot 18 + 54]

= \frac{27}{4 ^{10}} [405 + 72 + 2] = \frac{27 \cdot 479}{4 ^{10}}

Therefore,

k = 479

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Probability

Solution

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