Consider a coin that is weighted so that it is twice as likely to come up heads as tails. Answer the following questions using the Gaussian distribution description for the results of a series of coin tosses:a. What is the most likely outcome of a series of 2000 coin tosses using this biased coin?b. What is the relative probability of getting 1300 heads versus 1350 heads in a series of 2000 coin tosses using this coin?

0.754

Step-by-step explanation:

Given that, it is twice as likely to come up heads as tails in a coin ;

Let tail = T ; head = H

Total possible outcomes in the coin toss = 3

Number of heads = 2

Number of tails = 1

Probability = required outcome / Total possible outcomes

Then

P(T) = 1 / 3

P(H) = 2/3

What is the most likely outcome of a series of 2000 coin tosses using this biased coin?

Most likely outcome :

P = 2/3 = 0.667 ; n = 3

Using binomial distribution formula:

nCr * p^r * (1-p)^(n-r)

P=0 :

3C0 * 0.667^0 * (1 - 0.667)^(3-0) = 0.037

P=1 :

3C1 * 0.667^1 * (1 - 0.667)^(3-1) = 0.2218 = 0.222

P= 2 :

3C2 * 0.667^2 * (1 - 0.667)^(3-2)

= 0.444

b. What is the relative probability of getting 1300 heads versus 1350 heads in a series of 2000 coin tosses using this coin?

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following