Afair coin is tossed 16 times. what is the probability that exactly 7 tails occur?

To get the probability of obtaining exactly 7 heads if a coin is tossed 16 times, we model this using the binomial distribution;
p(x)=(n!)/[(n-x)!x!]p^xq^(n-x)
where:
n=number of ways an event can occur=16
x=number of times tails can occur=7
p=probability of success=1/2
q=probability of failure=1/2
thus the answer to our question will be:
p(x)=(16!)/[(16-7)!(7!])*(0.5)^7*(0.5)^(16-7)
=11,440*0.5^7*0.5^9
=0.17456
the answer is, the probability is 0.17456

Expression gives the perimeter = 4x√17

Step-by-step explanation:

∵ The diagonals of the rhombus ⊥ to each other and

  bisects each other

∴ Half the longest diagonal = 8x/2 = 4x

∴ Half the shortest diagonal = 2x/2 = x

Use Pythagoras theorem to find the length of the side of the rhombus

∴ The length of the side = √[(4x)² + (x)²] = √[16x² + x²] = √(17x²)

∴ S = x√17

∵ Perimeter of the rhombus = 4 × S

∴ P = 4 × x√17 = 4x√17