A table where every probability is listed: Picture a spinner with a few labelled outcomes. A discrete random variable X lists each value it can take and the probability of each.
Two facts to use
- The probabilities are a complete set of slices of one whole pie, so they must add to 1. That single fact lets you find any missing probability.
- Once the table is complete, the mean (also called the expected value) is each value weighted by how likely it is.
IB-style question — find the missing probability, then E(X)
A discrete random variable X has the probability table P(X=1)=0.2, P(X=2)=0.3, P(X=3)=k, P(X=4)=0.1.
Find the value of k, then find E(X).
Step by step
- The probabilities must add to 1. Add the known ones and subtract from 1.
- Solve for k.
- Now apply the mean formula: each value times its probability.
- Work out each product and add.
Final answer
k = 0.4 and E(X) = 2.4.
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Spread = 'mean of the squares' minus 'square of the mean': The mean tells you the centre; the variance tells you how spread out the values are.
The quick formula: E(X²) − [E(X)]²
- To get E(X²), square each value first, then weight by its probability — exactly like E(X) but with x² in place of x.
- Why subtract [E(X)]²? Because the squared values are inflated by where the centre sits.
- Removing the square of the mean leaves only the genuine spread about that centre.
IB-style question — variance from a table
X has P(X=1)=0.2, P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.1, so E(X)=2.4 from before.
Find Var(X).
Step by step
- First find E(X²): square each value, weight by its probability.
- Work out each term and add.
- Subtract the square of the mean.
- Compute (2.4)² = 5.76 and subtract.
Final answer
Var(X) = 0.84.