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:
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 is 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.