Question

Find the probability that ‘\[10\]’ is followed by ‘\[01\]’ given a computer game which generate only the digits from \[0\] to \[1\] to form a string of binary numbers with a probability of occurrence of \[0\] at even places be \[\dfrac{1}{2}\] and probability of occurrence of \[0\] at the odd place be \[\dfrac{1}{3}\].
A. \[\dfrac{1}{6}\]
B. \[\dfrac{1}{{18}}\]
C. \[\dfrac{1}{9}\]
D. \[\dfrac{1}{3}\]

Answer 1

Hint: Collect the probabilities of the events from the given information in the question and then apply total probability theorem to get required answer.

Formula used:
The probability that the event does not occur is \[=1-p\].

Complete step by step solution
We have, given in the question:
Probability ‘\[P\]’ of ‘\[0\]’ at even places:
\[P\] ( \[0\] at even places) = \[\dfrac{1}{2}\]
Probability ‘\[P\]’ of ‘\[0\]’ at odd places:
\[P\]( \[0\] at odd places) = \[\dfrac{1}{3}\]
Similarly,
\[P\]( \[1\] at even places) = \[\dfrac{1}{2}\]
\[P\]( \[1\] at odd places) = \[\dfrac{2}{3}\]
There are two possibilities to occur the event:

1	0	0	1
Odd	Even	Odd	Even

 OR

1	0	0	1
Even	Odd	Even	Odd

According above two cases:
\[P\]( \[10\] is followed by \[01\]) \[\left[ {\left( {\dfrac{2}{3}} \right) \times \left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{1}{3}} \right) \times \left( {\dfrac{1}{2}} \right)} \right] + \left[ {\left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{1}{3}} \right) \times \left( {\dfrac{1}{2}} \right) \times \left( {\dfrac{2}{3}} \right)} \right]\]
\[ = \left( {\dfrac{1}{{18}}} \right) + \left( {\dfrac{1}{{18}}} \right)\]
 \[ = \dfrac{1}{9}\]
The correct answer is option C.

Note: Students often do a common mistake to find the required probability. The strings are made by only 0 and 1. The string should be in the form 10 followed by 01. The string will be 1001. Don’t take only the case:

1	0	0	1
Even	Odd	Even	Odd

Since it is a string the position of 1 may an odd place. Students often donot consider the second case. The second case is:

1	0	0	1
Odd	Even	Odd	Even

So there are two possible cases.