Answer:
Some details are missing
Step-by-step explanation:
An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails) (t) which we write hth, ttt, etc. For each outcome, let R be the random variable counting the number of heads in each outcome. For example, if the outcome is hht, then R(hht) = 2. Suppose that the random variable X is defined in terms of R as follows: X = 2R² - 6R - 1. The values of X are thus:
Outcome: || Value of X
tht || -5
thh || -5
hth || -5
htt || -5
hhh || -1
tth || -5
hht || -5
ttt || -1
Calculate the probability distribution function of X, i.e. the function Px (x). First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.
Solution
To calculate the probability distribution function of X.
We have to observe the total outcomes to check the number of "Heads (h) " in each outcome.
The first, fourth and, sixth outcome has 1 head (h)
The second, third and seventh outcome has 2 head (hh)
The fifth outcome has 3 head (hhh)
The eight outcome has 0 appearance of h
We then solve the probability of each occurrence
i.e. The probability of having h, hh, hhh and no occurrence of h
This will be represented as follows
P(h=0)
P(h=1)
P(h=2)
P(h=3)
In a coin, the probability of getting a head = ½ and the probability of getting a tail = ½ in 1 toss
Using the following formula
P(X=x) = nCr a^r * b ^ (n-r)
Where n represents total number of toss = 3
r represents number of occurrence
a represents getting a head = ½
b represents probability of getting a tail = ½
1. For h = 0
P(h=0) = 3C0 * ½^0 * ½³
P(h=0) = 1 * 1 * ⅛
P(h=0) = ⅛
2. For h = 1
P(h=1) = 3C1 * ½^1 * ½²
P(h=1) = 3 * ½ * ¼
P(h=1) = ⅜
3. P(h=2) = 3C2 * ½² * ½^1
P(h=2) = 3 * ¼ * ½
P(h=2) = ⅜
4.P(h=3) = 3C3 * ½³ * ½^0
P(h=0) = 1 * ⅛ * 1
P(h=0) = ⅛
It should be noted that when X is -5, h is either 1 or 2 and P(X) = ⅜
When X is -1, h is either 0 or 3 and P(X) = ⅛
The probability distribution function of X is as follows
Values of X || P(x)
-5 || ⅜
1 || ⅛
or
