Question

A) What is the probability that when using an unfair coin, with a head = 1/5 and a tail =4/5:

Answer 1

Answer:

A)

1) 0.03072 = 3.072% probability you get the third head on the 5th toss.

2) The expected number of heads you will get when you toss the coin six times is 1.2.

B)

1) 0.2304 = 23.04% probability that you will get 2 heads and 3 tails.

2) The expected number of heads you will get when you toss the coin six times is 3.6.

Step-by-step explanation:

For each coin, there are only two possible outcomes. Either it is tails, or it is heads. The probability of a toss being heads or tails is independent of any other toss. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is E(X) = np.

Item A:

1) you get the third head on the 5th toss?

1/5 probability of heads, so $p = \frac{1}{5} = 0.2$

This is two heads on 4 tosses(P(X = 2) when n = 4) and then head on the 5th toss, with 0.2 probability. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.2)^{2}.(0.8)^{2} = 0.1536$

0.1536*0.2 = 0.03072

0.03072 = 3.072% probability you get the third head on the 5th toss.

2) that you get the expected number of heads you will get when you toss the coin six times?

Six tosses means that $n = 6$. So

$E(X) = np = 6*0.2 = 1.2$

The expected number of heads you will get when you toss the coin six times is 1.2.

Question B:

3/5 probability of heads, which means that $p = \frac{3}{5} = 0.6$

1) If you toss the coin 5 times, what is the probability that you will get 2heads and 3 tails?

This is P(X = 2) when n = 5. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{5,2}.(0.6)^{2}.(0.4)^{3} = 0.2304$

0.2304 = 23.04% probability that you will get 2 heads and 3 tails.

2) What is the expected number of heads you will get when you toss the coin six times?

Six times means that $n = 6$. So

$E(X) = np = 6(0.6) = 3.6$

The expected number of heads you will get when you toss the coin six times is 3.6.