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Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
We have 3 coins being tossed. Each coin has 2 possible outcomes: Heads (H) or Tails (T).
With 3 coins, there are 2 × 2 × 2 = 8 possible outcomes:
The complete sample space is:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Let X be the number of heads. We count heads in each outcome:
| Outcome | Number of Heads (X) |
|---|---|
| HHH | 3 |
| HHT, HTH, THH | 2 |
| HTT, THT, TTH | 1 |
| TTT | 0 |
The PMF gives the probability for each possible value of X:
| X (Number of Heads) | Number of Outcomes | P(X) |
|---|---|---|
| 0 | 1 | 1/8 = 0.125 |
| 1 | 3 | 3/8 = 0.375 |
| 2 | 3 | 3/8 = 0.375 |
| 3 | 1 | 1/8 = 0.125 |
The probability mass function is:
A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find:
(i) the probability mass function (ii) the cumulative distribution function (iii) P(4 ≤ X < 10) (iv) P(X ≥ 6)
First, let's understand the probabilities for one die:
| Face Value | Number of Faces | Probability |
|---|---|---|
| 1 | 1 | 1/6 ≈ 0.1667 |
| 3 | 2 | 2/6 ≈ 0.3333 |
| 5 | 3 | 3/6 = 0.5 |
When thrown twice, the possible combinations are:
| First Throw | Second Throw | Total (X) | Probability |
|---|---|---|---|
| 1 | 1 | 2 | (1/6)×(1/6) = 1/36 |
| 1 | 3 | 4 | (1/6)×(2/6) = 2/36 |
| 1 | 5 | 6 | (1/6)×(3/6) = 3/36 |
| 3 | 1 | 4 | (2/6)×(1/6) = 2/36 |
| 3 | 3 | 6 | (2/6)×(2/6) = 4/36 |
| 3 | 5 | 8 | (2/6)×(3/6) = 6/36 |
| 5 | 1 | 6 | (3/6)×(1/6) = 3/36 |
| 5 | 3 | 8 | (3/6)×(2/6) = 6/36 |
| 5 | 5 | 10 | (3/6)×(3/6) = 9/36 |
Now we can find the PMF by summing probabilities for each X:
| X (Total) | Outcomes | P(X) |
|---|---|---|
| 2 | (1,1) | 1/36 ≈ 0.0278 |
| 4 | (1,3), (3,1) | 4/36 ≈ 0.1111 |
| 6 | (1,5), (3,3), (5,1) | 10/36 ≈ 0.2778 |
| 8 | (3,5), (5,3) | 12/36 ≈ 0.3333 |
| 10 | (5,5) | 9/36 = 0.25 |
The CDF F(x) = P(X ≤ x):
| X | F(X) |
|---|---|
| X < 2 | 0 |
| 2 ≤ X < 4 | 1/36 ≈ 0.0278 |
| 4 ≤ X < 6 | 5/36 ≈ 0.1389 |
| 6 ≤ X < 8 | 15/36 ≈ 0.4167 |
| 8 ≤ X < 10 | 27/36 = 0.75 |
| X ≥ 10 | 36/36 = 1 |
This is F(9) - F(4) + P(X=4) = F(9) - F(3)
From the CDF table:
F(9) = 27/36 (for 8 ≤ X < 10)
F(3) = 1/36 (for 2 ≤ X < 4)
So P(4 ≤ X < 10) = 27/36 - 1/36 = 26/36 ≈ 0.7222
Alternatively, sum the probabilities for X=4,6,8:
4/36 + 10/36 + 12/36 = 26/36 ≈ 0.7222
This is 1 - F(5) = 1 - 5/36 = 31/36 ≈ 0.8611
Alternatively, sum probabilities for X=6,8,10:
10/36 + 12/36 + 9/36 = 31/36 ≈ 0.8611
Find the probability mass function and cumulative distribution function of number of girl child in families with 4 children, assuming equal probabilities for boys and girls.
Each child has equal probability (0.5) of being a girl (G) or boy (B). We're considering families with 4 children.
There are 2⁴ = 16 possible gender sequences for 4 children.
Let X be the number of girls in a family. X can be 0, 1, 2, 3, or 4.
This is a binomial distribution with n=4, p=0.5.
The PMF is given by:
Where C(4,k) is the number of combinations of 4 children taken k at a time.
| X (Girls) | Number of Outcomes | P(X) |
|---|---|---|
| 0 | 1 (BBBB) | 1/16 = 0.0625 |
| 1 | 4 (GBBB, BGBB, BBGB, BBBG) | 4/16 = 0.25 |
| 2 | 6 (GGBB, GBGB, GBBG, BGGB, BGBG, BBGG) | 6/16 = 0.375 |
| 3 | 4 (GGGB, GGBG, GBGG, BGGG) | 4/16 = 0.25 |
| 4 | 1 (GGGG) | 1/16 = 0.0625 |
The CDF F(x) = P(X ≤ x):
| X | F(X) |
|---|---|
| X < 0 | 0 |
| 0 ≤ X < 1 | 1/16 = 0.0625 |
| 1 ≤ X < 2 | 5/16 ≈ 0.3125 |
| 2 ≤ X < 3 | 11/16 ≈ 0.6875 |
| 3 ≤ X < 4 | 15/16 ≈ 0.9375 |
| X ≥ 4 | 16/16 = 1 |
Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by:
Find (i) the value of k (ii) cumulative distribution function (iii) P(X ≥ 1).
For a PMF, the sum of all probabilities must equal 1:
Now we can find each probability with k=8:
| x | f(x) |
|---|---|
| 0 | (0 + 1)/8 = 1/8 = 0.125 |
| 1 | (1 + 1)/8 = 2/8 = 0.25 |
| 2 | (4 + 1)/8 = 5/8 = 0.625 |
The CDF F(x) = P(X ≤ x):
| x | F(x) |
|---|---|
| x < 0 | 0 |
| 0 ≤ x < 1 | 1/8 = 0.125 |
| 1 ≤ x < 2 | 3/8 = 0.375 |
| x ≥ 2 | 8/8 = 1 |
This is 1 - P(X < 1) = 1 - F(0) = 1 - 1/8 = 7/8 = 0.875
Alternatively, sum probabilities for X=1 and X=2:
2/8 + 5/8 = 7/8 = 0.875
The cumulative distribution function of a discrete random variable is given by:
Find (i) the probability mass function (ii) P(X < 1) and (iii) P(X ≥ 2).
The PMF is the difference in CDF at jump points:
| x | P(X = x) = F(x) - F(x⁻) |
|---|---|
| -1 | 0.15 - 0 = 0.15 |
| 0 | 0.35 - 0.15 = 0.20 |
| 1 | 0.60 - 0.35 = 0.25 |
| 2 | 0.85 - 0.60 = 0.25 |
| 3 | 1.00 - 0.85 = 0.15 |
This is F(1⁻) = 0.35
This is 1 - F(2⁻) = 1 - 0.60 = 0.40
Alternatively, sum probabilities for X=2 and X=3:
0.25 + 0.15 = 0.40
A random variable X has the following probability mass function:
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| f(x) | k² | 2k² | 3k² | 2k | 3k |
Find (i) the value of k (ii) P(2 ≤ X < 5) (iii) P(3 < X)
The sum of all probabilities must equal 1:
Solving the quadratic equation:
This includes X=2,3,4:
This includes X=4,5:
The cumulative distribution function of a discrete random variable is given by:
Find (i) the probability mass function (ii) P(X < 3) and (iii) P(X ≥ 2).
First, note the CDF values should be probabilities between 0 and 1. Assuming they're multiplied by some factor (likely 10):
Divide all values by 10 to get proper probabilities:
| x | P(X = x) = F(x)/10 - F(x⁻)/10 |
|---|---|
| 0 | 0.1 - 0 = 0.1 |
| 1 | 0.2 - 0.1 = 0.1 |
| 2 | 0.5 - 0.3 = 0.2 |
| 3 | 0.9 - 0.5 = 0.4 |
| 4 | 1.0 - 0.9 = 0.1 |
This is F(3⁻)/10 = 0.5
This is 1 - F(2⁻)/10 = 1 - 0.3 = 0.7
Alternatively, sum probabilities for X=2,3,4:
0.2 + 0.4 + 0.1 = 0.7
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