Question

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Q‒2. [5×4 marks] Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that
a) the bit string has exactly two 1s;
b) the bit string begins and ends with 0;
c) the bit string has the sum of its digits equal to seven;
d) the bit string has more 0s than 1s;
e) the bit string has exactly two 1s, given that the string begins with a 1.

Answer 1

a. 45/1024

b. 1/4

c. 15/128

d. 193/512

e. 9/256

Explanation:

Here, each position can be either a 0 or a 1.

So, total number of strings possible = 2^10 = 1024

a) For strings that have exactly two 1's, it means there must also be exactly eight 0's.

Thus, total number of such strings possible

10!/2!8!=45

Thus, probability

45/1024

b) Here, we have fixed the 1st and the last positions, and eight positions are available.

Each of these 8 positions can take either a 0 or a 1.

Thus, total number of such strings possible

=2^8=256

Thus, probability

256/1024 = 1/4

c) For sum of bits to be equal to seven, we must have exactly seven 1's in the string. Also, it means there must also be exactly three 0's

Thus, total number of such strings possible

10!/7!3!=120

Thus, probability

120/1024 = 15/128

d) Following are the possibilities :

There are six 0's, four 1's :

So, number of strings

10!/6!4!=210

There are seven 0's, three 1's :

So, number of strings

10!/7!3!=120

There are eight 0's, two 1's :

So, number of strings

10!/8!2!=45

There are nine 0's, one 1's :

So, number of strings

10!/9!1!=10

There are ten 0's, zero 1's :

So, number of strings

10!/10!0!=1

Thus, total number of string possible

= 210 + 120 + 45 + 10 + 1

= 386

Thus, probability

386/1024 = 193/512

e) Here, we have fixed the starting position, so 9 positions remain.

In these 9 positions, there must be exactly two 1's, which means there must also be exactly seven 0's.

Thus, total number of such strings possible

9!/2!7!=36

Thus, probability

36/1024 = 9/256