AQA · Mathematics · 8300
GCSE Maths: Probability: conditional
Probability · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Probability: conditional (Corbettmaths Video 247), and every card links back to the exact page it came from. Practice tests are drawn from this deck’s own cards. Source material © Corbettmaths — used with attribution.
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In conditional probability problems, what does "without replacement" mean?
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Once an item is selected, it is not returned to the collection before the next selection, so the total number of items decreases.
When selecting items without replacement, what happens to the denominator in probability calculations for the second selection?
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The denominator decreases by 1 because there is one fewer item in the total collection.
For two independent events A and B, how do you calculate P(A and B)?
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Multiply the probabilities: P(A and B) = P(A) × P(B)
When calculating the probability of "at least one" event occurring, what complementary approach can simplify the calculation?
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Calculate P(none) and subtract from 1: P(at least one) = 1 - P(none)
In a two-stage probability problem where the first event determines which path to take, what is this type of problem called?
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A tree diagram or conditional probability problem with multiple branches/stages.
If you want the probability of selecting two items of the same type from a mixed collection, what general approach do you use?
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Calculate P(both type A) + P(both type B) + ... for each type, then add the probabilities together.
When calculating conditional probability with replacement vs. without replacement, which gives a lower probability for selecting the same type twice?
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Without replacement gives lower probability because the second selection has fewer items of the desired type.
In conditional probability, what does P(A|B) represent?
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The probability of event A occurring given that event B has already occurred.
When using a tree diagram for multi-stage probability, how do you find the probability of a specific outcome path?
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Multiply the probabilities along the branches that lead to that outcome.
If there are n items of a specific type, what is the probability of picking two of that type without replacement?
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(n/(total)) × ((n-1)/(total-1)), where the numerator and denominator both decrease by 1 for the second pick.
In problem 26, if there are x apples with 4 bad apples, what is the probability of selecting two bad apples without replacement?
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(4/x) × (3/(x-1)), which simplifies to 12/(x(x-1)).
If the probability of selecting two bad apples (4 out of x total) without replacement equals 1/11, what equation results?
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12/(x(x-1)) = 1/11, which cross-multiplies to 12 × 11 = x(x-1), giving x² - x - 132 = 0.
When a ratio of counters is given as 2:2:1, how do you express the number of each type in terms of a variable?
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Let k be the multiplier, so the quantities are 2k, 2k, and k respectively. The total is 5k.
In problem 25, if yellow counters number 2k out of 5k total, what is the probability of drawing two yellow counters without replacement?
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(2k/5k) × ((2k-1)/(5k-1)), which simplifies to (2k(2k-1))/(5k(5k-1)).
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