Math · Introductory statistics · Worked example

Find the probability of at least one success

Each ticket in a raffle wins a prize with probability $0.1$ , independently of the others. You hold $5$ tickets. What is the probability that at least one wins, and how many wins do you expect?

Name the model

$X$ , the number of winning tickets, counts successes in $n = 5$ independent trials with $p = 0.1$ .

X \sim B (5, 0.1)

Use the complement

“At least one win” covers $X = 1$ , $2$ , $3$ , $4$ and $5$ . Its opposite is the single case $X = 0$ , so find that and subtract it from $1$ .

P (X \geq 1) = 1 - P (X = 0)

Find $P (X = 0)$ and subtract

No wins means five losing tickets, each with probability $0.9$ , and independent probabilities multiply: ${(0.9)}^{5} = 0.59049$ .

\begin{array}{l} X \sim B (5, 0.1) \\ P (X \geq 1) = 1 - P (X = 0) \\ = 1 - {(0.9)}^{5} \\ = 1 - 0.59049 \\ = 0.40951 \end{array}

Compare with the expected value

$E (X) = n p = 0.5$ . Half a win on average is not a $50 %$ chance of winning: the probability of at least one win is about $0.41$ . They differ because some outcomes hold two or more wins, and those count more than once in the average.

\begin{array}{l} X \sim B (5, 0.1) \\ E (X) = n p = 5 \cdot 0.1 = 0.5 \end{array}

Result

$P (X \geq 1) = 0.40951$ , about $0.41$ , and $E (X) = 0.5$ wins.

Your turn

1. One like the example

A fair die is rolled $4$ times. What is the probability of at least one six?

Show the answer and explanation

$671 / 1296 \approx 0.518$ .

The chance of no six on one roll is $5 / 6$ , so the chance of no six in four rolls is ${(5 / 6)}^{4} = 625 / 1296$ . Subtract from $1$ to get $671 / 1296 \approx 0.518$ . Chalk reads $B (4, 1 / 6)$ written with a slash.

\begin{array}{l} X \sim B (4, 1 / 6) \\ P (X \geq 1) = 1 - P (X = 0) \\ = 1 - {(\frac{5}{6})}^{4} \\ = \frac{671}{1296} \\ \frac{671}{1296} \approx 0.518 \end{array}

2. A variation

Each of $6$ light bulbs fails within a year with probability $0.1$ , independently. What is the probability that at most one fails?

Show the answer and explanation

$0.885735$ , about $0.886$ .

“At most one” is $X = 0$ or $X = 1$ , two terms, so add them directly. $P (X = 0)$ $= {0.9}^{6}$ $= 0.531441$ and $P (X = 1)$ $= 6 (0.1) {(0.9)}^{5}$ $= 0.354294$ .

\begin{array}{l} X \sim B (6, 0.1) \\ P (X \leq 1) = P (X = 0) + P (X = 1) \\ = {(0.9)}^{6} + 6 (0.1) {(0.9)}^{5} \\ = 0.531441 + 0.354294 \\ = 0.885735 \end{array}

3. Somewhere new

Each raffle ticket wins with probability $0.1$ , independently. How many tickets must you hold for at least a $90 %$ chance of at least one win?

Show the answer and explanation

$22$ tickets.

With $n$ tickets, $P (at least one win)$ $= 1 - {0.9}^{n}$ . Set it equal to $0.9$ and solve, ${0.9}^{n} = 0.1$ , so $n = ln 0.1 / ln 0.9$ $\approx 21.85$ . A count must be whole, and $21$ tickets give only $0.8906$ , so round up. $22$ tickets give $0.9015$ .

\begin{array}{l} 1 - {0.9}^{n} = 0.9 \\ {0.9}^{n} = 0.1 \\ n = \frac{ln 0.1}{ln 0.9} \\ n \approx 21.85 \end{array}

Keep exploring

In the Math box, change the second row to $P (X \geq 1) = 1 - P (X = 1)$ , a common slip. The row turns to ✗: leaving out one win is not the opposite of at least one win.

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Authored study material. Tool results depend on the stated inputs and model assumptions.

After reading

Study it, then try it

Practice the key terms as flashcards, then open the example in its tool, change a value and keep a useful result on your board.

Study the 7 key terms Check the complement in Math Drop balls through 5 rows in Binomial & the Galton Board Open worked example on a board Binomial probability in Math Reference

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Name the model

Use the complement

Find P⁢(X=0)P(X=0) and subtract

Compare with the expected value

Result

Your turn

1. One like the example

671⁢/1296≈0.518671/1296≈0.518.

2. A variation

0.8857350.885735, about 0.8860.886.

3. Somewhere new

2222 tickets.

Keep exploring

Study it, then try it

Find $P (X = 0)$ and subtract

$671 / 1296 \approx 0.518$ .

$0.885735$ , about $0.886$ .

$22$ tickets.