What is the probability that e) A fair coin lands Heads 6 times in a row? f) A fair coin lands Heads 4 times out of 5 flips? g) The 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? 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?

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%

whatvare the choices

step-by-step explanation:

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