JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 2)
Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $${1 \over 2}$$ and probability of occurrence of 0 at the odd place be $${1 \over 3}$$. Then the probability that '10' is followed by '01' is equal to :
$${1 \over 18}$$
$${1 \over 3}$$
$${1 \over 9}$$
$${1 \over 6}$$
Explanation
P(0 at even place) = $${1 \over 2}$$
P(0 at odd place) = $${1 \over 3}$$
P(1 at even place) = $${1 \over 2}$$
P(1 at odd place) = $${2 \over 3}$$
P(10 is followed by 01)
= $$\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 3} \times {1 \over 2} \times {2 \over 3}} \right)$$
= $${1 \over {18}} + {1 \over {18}}$$
= $${1 \over 9}$$
P(0 at odd place) = $${1 \over 3}$$
P(1 at even place) = $${1 \over 2}$$
P(1 at odd place) = $${2 \over 3}$$
P(10 is followed by 01)
= $$\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 3} \times {1 \over 2} \times {2 \over 3}} \right)$$
= $${1 \over {18}} + {1 \over {18}}$$
= $${1 \over 9}$$
Comments (0)
