An urn holds 5 red balls and 3 blue. Two are drawn without replacement. What is the probability both are red?

On the first draw: 5 out of 8. After it, 4 reds remain out of 7 balls. Multiply: five eighths times four sevenths. 20 over 56, about 0.357

From the same urn, two are drawn with replacement. What is the probability both are red?

The urn returns to its original state after the first draw. Both draws have the same probability: 5 out of 8. Multiply: five eighths squared. 25 over 64, about 0.391 — higher than without replacement

A shooter hits 70 percent of attempts. What is the probability of exactly 3 hits out of 5?

The number of ways to choose 3 out of 5 is 10. Success to the power 3: 0.7 cubed, which is 0.343. Failure to the power 2: 0.3 squared, which is 0.09. Multiply the three factors. About 0.309

Mathematics, four units · Grade 11

What changes in the calculation when you draw without replacement?

The denominator changes, and sometimes the numerator too. With replacement the situation resets and every draw is independent, so you multiply the same probability. Without replacement, once one ball is out there are fewer balls in total and fewer of the colour drawn, so the second probability is worked out from a new situation.

Learning objectives

Organise the data of a probability question into a two way table
Calculate a probability for a draw with replacement and for one without
Apply the binomial model to a fixed number of repeated trials
Decide whether repeated trials in a question are genuinely independent

The two-way table

Most probability questions on the paper describe a population split by two criteria — boys and girls against passes and failures, say. The right tool is a table with two rows, two columns, and a row and column of totals.

Fill in what is given and complete the rest by subtraction. In most questions three numbers are given and everything else follows, and anyone who fills the table right through finds the answer sitting inside it.

Conditional probability is read straight off the table: the probability a pupil passed given that they are a boy is the number of boys who passed over the total number of boys — you divide by the row, not by the grand total.

With and without replacement

With replacement, the item goes back before the next draw. The situation is identical every time, the probabilities do not change, and the events are independent.

Without replacement every draw alters the pool. With 5 red and 3 blue in an urn, the probability of two reds is five eighths times four sevenths — both the numerator and the denominator have dropped by one.

A question about "at least one" is almost always easier through the complement: one minus the probability of none. Computing "at least one" directly means adding several cases, most of them unnecessary.

The binomial experiment

The binomial model describes a fixed number of repeated trials, each with exactly two outcomes — success or failure — where the probability of success is the same every time and the trials are independent.

The probability of exactly k successes out of n is the number of ways to choose which k succeeded, times the probability of success to the power k, times the probability of failure to the power n minus k. All three factors are always needed.

Check the conditions before applying it. Drawing without replacement is not binomial, because the probability changes from trial to trial. That distinction is exactly what this item is testing, so it is worth marks in its own right.

Worked examples

An urn holds 5 red balls and 3 blue. Two are drawn without replacement. What is the probability both are red?
1. On the first draw: 5 out of 8
2. After it, 4 reds remain out of 7 balls
3. Multiply: five eighths times four sevenths
Answer: 20 over 56, about 0.357
From the same urn, two are drawn with replacement. What is the probability both are red?
1. The urn returns to its original state after the first draw
2. Both draws have the same probability: 5 out of 8
3. Multiply: five eighths squared
Answer: 25 over 64, about 0.391 — higher than without replacement
A shooter hits 70 percent of attempts. What is the probability of exactly 3 hits out of 5?
1. The number of ways to choose 3 out of 5 is 10
2. Success to the power 3: 0.7 cubed, which is 0.343
3. Failure to the power 2: 0.3 squared, which is 0.09
4. Multiply the three factors
Answer: About 0.309

Common mistakes

Keeping the same denominator when drawing without replacement: After one draw there is one fewer item in total. Reducing the numerator alone is a half-correction that produces a consistently wrong answer.
Applying the binomial model to drawing without replacement: The binomial model requires a constant probability and independence. Drawing without replacement breaks both, and the right model there is a product of conditional probabilities.
Computing "at least one" by adding cases: It is possible but long, and a case usually goes missing on the way. One minus the probability of none gives the same answer in a single line with nothing left out.

What to remember

A two-way table solves most of these questions.
Conditional: divide by the row, not the grand total.
Without replacement: both numerator and denominator drop.
At least one equals one minus none.