JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 8)
Four fair dice are thrown independently 27 times. Then the expected number of times, at
least two dice show up a three or a five, is _________.
Answer
11
Explanation
4 dice are independently thrown. Each die has probability to show 3 or 5 is
$$P = {2 \over 6} = {1 \over 3}$$
$$ \therefore $$ $$q = 1 - {1 \over 3} = {2 \over 3}$$ (not showing 3 or 5)
Experiment is performed with 4 dices independently
$$ \therefore $$ Their binomial distribution is
$${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$$
$$ \therefore $$ In one throw of each dice probability of showing 3 or 5 at least twice is
= $${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$$
$$ = {{33} \over {81}}$$
Given such experiment performed 27 times
$$ \therefore $$ So expected outcomes = np
= $${{33} \over {81}} \times 27$$
= 11
$$P = {2 \over 6} = {1 \over 3}$$
$$ \therefore $$ $$q = 1 - {1 \over 3} = {2 \over 3}$$ (not showing 3 or 5)
Experiment is performed with 4 dices independently
$$ \therefore $$ Their binomial distribution is
$${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$$
$$ \therefore $$ In one throw of each dice probability of showing 3 or 5 at least twice is
= $${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$$
$$ = {{33} \over {81}}$$
Given such experiment performed 27 times
$$ \therefore $$ So expected outcomes = np
= $${{33} \over {81}} \times 27$$
= 11
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