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A second set of probability equation questions questions for GCSE Maths. They are different from the ones on the topic page, so work through both. Try each question, then open the answer to check your working.
A bag has n counters, 6 of them red. Two are taken with replacement. P(both red) = 1/4. Find n.
Answer: 12
The equation n^2 - n - 12 = 0 comes from a bag. The roots are 4 and -3. Which is the number of counters?
Answer: 4
n marbles, 7 red. P(both red, no replacement) = 1/5. Find n.
Answer: 15
Roots of the cleared equation are n = 6 and n = -5. The bag started with 4 red counters. Which root is possible?
Answer: 6
n coins, 3 special. P(both special, no replacement) = 1/7. Find n.
Answer: 7
After clearing, n^2 - n - 90 = 0. The story is a bag of counters. Give the only possible n.
Answer: 10
A longer question with marks, in the style of a written paper. This is a Yumio practice question, not a past paper question.
A bag contains n counters, of which 6 are red. Two counters are taken at random without replacement. The probability that both are red is 1/3. Find n.
Answer: 10
Short one-step questions. Pick an option before you open the answer.
Why does the second denominator become n - 1?
One counter has already been taken out. One counter has already been taken out.
Why does the second numerator fall by 1 when you want two reds?
One red has already been removed. One red has already been removed.
What equation form do you reach from n(n-1) = 90?
n^2 - n - 90 = 0. n^2 - n - 90 = 0.
Why reject a negative root?
The bag cannot contain a negative number of counters. The bag cannot contain a negative number of counters.
P(both red) = 1/3 with 6 red counters leads to which n?
10. Check (6/10)*(5/9) = 1/3. 10. Check (6/10)*(5/9) = 1/3.
What is different about with replacement?
The second fraction would still have denominator n. The second fraction would still have denominator n.
3 mint out of n, both mint probability 1/2. What is n?
4 counters. Check (3/4) times (2/3). 4 counters. Check (3/4) times (2/3).
If a root is smaller than the number of red counters, why reject it?
You cannot draw more red counters than the bag holds. You cannot draw more red counters than the bag holds.
Which two fractions multiply for two draws?
The probability of the first outcome, then the probability of the second given the first. The probability of the first outcome, then the probability of the second given the first.
n(n-1) = 12 with a positive bag. What is n?
4, because 4 times 3 is 12. 4, because 4 times 3 is 12.
Name the step after you have the quadratic.
Solve it, then keep only the root that makes sense in the story. Solve it, then keep only the root that makes sense in the story.
(5/10)*(4/9) equals which fraction?
2/9. Multiply the tops and the bottoms to get 20/90, then simplify. 2/9. Multiply the tops and the bottoms to get 20/90, then simplify.
Why write a product rather than a sum?
Both draws happen, so you multiply along the path. You add only if you are combining different paths. Both draws happen, so you multiply along the path. You add only if you are combining different paths.
A root n = -3 appears. What do you do with it?
Discard it and give the positive root. Discard it and give the positive root.
What should you check at the end?
Put n back into the two fractions and see that you recover the probability in the question. Put n back into the two fractions and see that you recover the probability in the question.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.