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 :
Solution
P(0 at even place) = ${1 \over 2}$
<br><br>P(0 at odd place) = ${1 \over 3}$
<br><br> P(1 at even place) = ${1 \over 2}$
<br><br>P(1 at odd place) = ${2 \over 3}$
<br><br>P(10 is followed by 01)
<br><br>= $$\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 3} \times {1 \over 2} \times {2 \over 3}} \right)$$
<br><br>= ${1 \over {18}} + {1 \over {18}}$
<br><br>= ${1 \over 9}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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