Question

suppose you have an expirement where you toss a fair coin three times you the count the number of heads observed over those three tosses use this experiment to address each of the following questions round solution to three decimal places

Answer 1

The experiment is to toss a fair coin three times and count the number of heads observed over those three tosses.

(a) Random variable

Let X be the random variable.

X = the number of heads observed when you flip a coin three times.

(b) Probability distribution of x

Let us first find the sample space for this experiment.

When you flip a coin three times then the following outcomes are possible

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Count the number of times you get 0 heads (x = 0)

1 ( TTT)

So, the probability is

P(x=0) = 1/8 = 0.125

Count the number of times you get 1 head (x = 1)

3 (HTT, THT, TTH)

So, the probability is

P(x=1) = 3/8 = 0.375

Count the number of times you get 2 heads (x = 2)

3 (HHT, HTH, THH)

So, the probability is

P(x=2) = 3/8 = 0.375

Count the number of times you get 3 heads (x = 3)

1 (HHH)

So, the probability is

P(x=3) = 1/8 = 0.125

Therefore, the probability distribution table of x is

(c) Shape of probability distribution of x

The shape of the probability distribution of x seems to be normal (symmetric)

(d) mean number of heads

The mean or expected number of heads is given by

`\mu=\sum x\cdot p(x)`

Where x is the number of heads and p(x) is the corresponding probability.

`\begin{gathered} \mu=\sum x\cdot p(x) \\ \mu=0\cdot0.125+1\cdot0.375+2\cdot0.375+3\cdot0.125 \\ \mu=0+0.375+0.75+0.375 \\ \mu=1.5 \end{gathered}`

Therefore, the mean number of heads is 1.5

This means that on average, you are expected to get 1.5 heads.