Question 8
The table shows the probability distribution for a random variable $X$ in a Bernoulli experiment. The random variable has only two possible values: 0 represents failure and 1 represents success.
| $x$ | 0 | 1 |
|---|---|---|
| $P(X=x)$ | 0.6 | 0.4 |
If six Bernoulli experiments are conducted, determine the probability of getting exactly two successes.
Not quite. 0.138 is $P(X=4)$, four successes and two failures. You’ve swapped the probabilities. Success has probability 0.4, not 0.6. In 2025, 6.7% of Queensland students chose (A).
Not quite. $0.4^2=0.16$ is the chance of the two successes on their own. It leaves out the four failures and the 15 orders the results can come in. In 2025, 14.9% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2025, 73.5% of Queensland students chose (C).
Not quite. $0.6^2=0.36$ squares the failure probability. You need two successes at 0.4 each, four failures at 0.6 each, and the 15 orders they can come in. In 2025, 4.5% of Queensland students chose (D).
This is the question. Two successes and four failures can come in 15 orders, and each order has probability $0.4^2\times0.6^4$. Together that’s 0.311.
This is four successes, 0.138, option (A). It’s the same as exactly two failures, which isn’t what the question asks.
Here success and failure have swapped, so this is 0.138, option (A). The table gives success a probability of 0.4.
The seven bars always add to 1. As $p$ grows, the tallest bars move right, towards more successes.
Six Bernoulli trials, each a success with the same probability 0.4. Counting the successes gives a binomial random variable.
$$X\sim\text{Bin}(6,\,0.4)$$
Exactly two successes means four failures. Any one order, like SSFFFF, has probability
$$0.4^2\times0.6^4=0.020736$$
There are $\binom{6}{2}=15$ ways to choose which two trials are the successes.
Multiply the number of orders by the probability of each one.
$\displaystyle P(X=2)=\binom{6}{2}(0.4)^2(0.6)^4$$\displaystyle =15\times0.16\times0.1296$$\displaystyle \approx0.311$
On your calculator, use the binomial probability (pdf) function with $n=6$, $p=0.4$ and $x=2$.
$$X\sim\text{Bin}(6,\,0.4)$$
$\displaystyle P(X=2)=\binom{6}{2}(0.4)^2(0.6)^4$$\displaystyle =0.31104$$\displaystyle \approx0.311$
How Queensland students answered in 2025
From the QCAA Mathematical Methods subject report for 2025. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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