Say you have a biased coin and you associate obtaining a head with the value 0 whereas obtaining a tail has a value of 2. Let X be the sum of the values obtained from tossing the coin three times.

Question

(a) If the probability of getting heads in a single toss is 0.32, then what is the probability distribution of X ?

x	0	2	4	6
P(X = x)

(b) What is P(X < 6)?

(c) What is P(1 < X < 5)?

please help thank you so much

Answer 1

One way to solve this problem is with the binomial distribution probability density function (PDF). If you are allowed to use them, they should be in your text or you can find them online at https://en.wikipedia.org/wiki/Binomial_distribution

But more intuitively, you can set this up by listing every outcome and then calculating its probability:

If you toss a coin 3 times, there are only 8 (2^3) possible outcomes:

HHH, HHT, HTH, HTT, THH, THT, TTH. TTT

We need to know (1) what each combination "scores" under the scoring system, and (2) what the probability of each of the 8 outcomes is. So make a table. I'm going to talk through the first two entries, and leave the rest for you. For HHH, the rule says that H=0, so three H's have a score of 0. The probability of getting H is 0.32 (given), so the probability of getting 3 H's in a row is (0.32)x(0.32)x(0.32). By the same logic HHT since H's are worth zero and T's are worth 2, HHT scores 2, and if the probability of H is 0.32, T is its converse (1-P(H)) or 1 - 0.32 = 0.68. So the probability of HHT is (0.32)x(0.32)x(0.68)

Tosses Score Probability

HHH 0 (0.32)x(0.32)x(0.32)

HHT 2 (0.32)x(0.32)x(0.68)

HTH

HTT

THH

THT

TTH

TTT

I hope this helps you get started. With this table, you should be able to complete the table you need to create and answer the two questions you've been given.

Answer 2

We can use binomial probability formula to compute probabilities:

P(x successes in n tries) = C(n,x) p^x (1-p)^(n-x)

0 = probability of getting 3 heads in 3 tosses = 0.32 * 0.32 * 0.32 = 0.032768

2 = probability of getting two heads in 3 tosses = 0.32 * 0.32 * 0.68 * 3 = 0.208896

4 = probability of getting one head in 3 tosses = 0.32 * 0.68 * 0.68 * 3 = 0.443904

6 = probability of getting zero heads in 3 tosses = 0.68 * 0.68 * 0.68 = 0.314432

P(< 6) = 1 - 0.314432 = 0.685568

P(1 < x < 5) = P(2) + P(4) = 0.208896 + 0.443904 = 0.6528