Answer:
a. 45/1024
b. 1/4
c. 15/128
d. 193/512
e. 9/256
Step-by-step 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 is
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 is
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 is
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 is
36/1024 = 9/256
