#1**0 **

Hello

If a and b are real numbers and n is a positive integer, then

(a + b)

n

=nC0

a

n + nC1

a

n – 1 b

1

+ nC2

a

n – 2 b

2

+ ...

... + nCr

a

n – r

br

+ ... + nCn

b

n

, where nCr

=

n

r n r −

for 0 ≤ r ≤ n

The general term or (r + 1)th term in the expansion is given by

Tr + 1 = nCr

a

n–r b

r here you can find more dinar chronicles

reneau34Feb 21, 2023

#1**0 **

To solve this problem, we can start by listing all the possible outcomes of flipping a fair coin two times:

HH (two heads)

HT (one head, one tail)

TH (one tail, one head)

TT (two tails)

We can then calculate the length of the longest run for each outcome:

HH: The longest run is 2.

HT: The longest run is 1.

TH: The longest run is 1.

TT: The longest run is 2.

Therefore, the expected length of the longest run is the average of the longest runs for each outcome:

Expected longest run = (2 + 1 + 1 + 2) / 4 = 1.5

So the expected length of the longest run when flipping a fair coin two times is 1.5.

reneau34Feb 20, 2023