find the missing probability in the table below, assume that X can only be 0
X P(X)
0 0.01
1 0.15
2 0.25
3 0.10
4 0.20
5 ???
a.)Given the rest of the table, what is P(X =5)?
b) What is P(X=2)?
c) What is P(X 3)?
d) What is E(X)?
The Correct Answer and Explanation is :
Let’s solve the missing probability and answer the related probability questions.
Given Probability Table:
| X | P(X) |
|---|---|
| 0 | 0.01 |
| 1 | 0.15 |
| 2 | 0.25 |
| 3 | 0.10 |
| 4 | 0.20 |
| 5 | ??? |
(a) Find ( P(X = 5) ):
Since the sum of all probabilities in a probability distribution must equal 1:
[
P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1
]
Substituting the known values:
[
0.01 + 0.15 + 0.25 + 0.10 + 0.20 + P(X=5) = 1
]
[
0.71 + P(X=5) = 1
]
[
P(X=5) = 1 – 0.71 = 0.29
]
(b) Find ( P(X=2) ):
From the table:
[
P(X=2) = 0.25
]
(c) Find ( P(X \geq 3) ):
[
P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)
]
[
P(X \geq 3) = 0.10 + 0.20 + 0.29
]
[
P(X \geq 3) = 0.59
]
(d) Compute the Expected Value ( E(X) ):
The expected value ( E(X) ) is calculated as:
[
E(X) = \sum X \cdot P(X)
]
[
E(X) = (0 \times 0.01) + (1 \times 0.15) + (2 \times 0.25) + (3 \times 0.10) + (4 \times 0.20) + (5 \times 0.29)
]
[
E(X) = 0 + 0.15 + 0.50 + 0.30 + 0.80 + 1.45
]
[
E(X) = 3.20
]
Explanation:
The missing probability was determined by ensuring the total probability equals 1, which is a fundamental rule in probability distributions. The probability that ( X=5 ) was calculated as 0.29.
For ( P(X=2) ), we directly referenced the table.
For ( P(X \geq 3) ), we summed the probabilities of all values of ( X ) that are at least 3.
The expected value, ( E(X) ), is essentially the weighted average of ( X ), considering the probabilities as weights. It provides an idea of the “center” of the distribution. In this case, ( E(X) = 3.20 ), meaning that, on average, the expected value of ( X ) is around 3.2.
This approach to solving probability distribution questions ensures an understanding of fundamental probability concepts like probability summation, cumulative probability, and expected values.