Answer:
e) 1.56%
f) 15.62%
h) 0.879%
g) 11.72%
Step-by-step explanation:
What we will do is solve point by point.
e) A fair coin lands Heads 6 times in a row?
We have the following:
Total number of possible outcomes = 2 ^ 6 = 64
Number of favorable outcomes = 1
Required probability = 1/64 = 1.56%
f) A fair coin lands Heads 4 times out of 5 flips
We have the following:
Total number of possible outcomes = 2 ^ 5 = 32
Number of favorable outcomes = 5C4
nCr = n! / (r! * (n-r)!)
5C4 = 5! / (4! * (5-4)!) = 5
Required probability = 5/32 = 15.62%
g) he bit string has exactly two 1s, given that the string begins with a 1 if you pick a bit string from the set of all bit strings of length ten?
We have the following:
Total number of possible outcomes = 2 ^ 10 = 1024
Number of ways in which a position excluding the start of the string can be chosen is 9C1
Total number of favorable outcomes = 9C1
9C1 = 9! / (1! * (9-1)!) = 9
Required probability = 9/1024 = 0.879%
h)The bit string has the sum of its digits equal to seven if you pick a bit string from the set of all bit strings of length ten?
We have the following:
Total number of possible outcomes = 2 ^ 10 = 1024
For the sum of the digits to be 7 there has to be 7 ones.
Number of ways in which 7 position can be chosen is 10C7.
Total number of favorable outcomes = 10C7
10C7 = 10! / (7! * (10-7)!) = 120
Required probability = 120/1024 = 11.72%
