The word ‘given’ changes the group you are counting within. Many probability errors come from using the original total after new information has ruled outcomes out. Draw the restricted group before choosing a denominator.
Curriculum checked 2 October 2026 · original practice for selected skills.
You will need: Fractions, equally likely outcomes and basic probability.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
P(A given B) is P(A and B) divided by P(B), provided P(B)>0. The denominator describes the information already given.
Independent events preserve probabilities when the other event is known. Mutually exclusive events with positive probabilities cannot be independent.
Reversing the condition usually changes the probability. Bayes reasoning includes both the likelihood of the observation and how common each source was before it.
Discard odd outcomes before counting.
The remaining outcomes are 2, 4 and 6.
Three equally likely outcomes remain.
Two of them, 4 and 6, exceed 3.
Answer: 2/3
Before moving on: can you explain why your method works, as well as give the answer?
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
The given event determines the denominator.
Divide 0.18 by 0.30.
P(B) is positive, so the conditional probability exists.
18/30 simplifies to 3/5.
Answer: 0.6
Before moving on: can you explain why your method works, as well as give the answer?
Compare the intersection with the product.
Compute 0.4 × 0.5.
The product is 0.2, equal to the intersection.
Knowing B does not change the probability of A.
Answer: Yes, they are independent.
Before moving on: can you explain why your method works, as well as give the answer?
The chance of red from bag A differs from the chance that a red ball came from A.
A contributes (1/4)(3/4)=3/16 to red; B contributes (3/4)(1/4)=3/16.
Given A, only its four balls matter. Three are red, so P(red given A)=3/4.
Multiply each bag's chance of being chosen by its chance of giving red, then add.
Among all red draws, A contributes 3/16 out of a total 3/8. The denominator is the chance of red from either bag.
Answer: P(red given A)=3/4; P(A given red)=1/2.
Before moving on: can you explain why your method works, as well as give the answer?
A tempting claim: “P(A given B) always equals P(B given A).”
The two conditions restrict the sample space differently.
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
First remove outcomes that do not fit the given condition.
The remaining outcomes are 4, 5 and 6.
Given a result greater than 3, only 4, 5 and 6 remain. These three outcomes are equally likely.
Only 5 is odd, so there is one success out of three remaining outcomes.
Answer: 1/3
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Use D in the denominator.
0.12/0.4.
The intersection is the part of D also in C.
12/40=3/10.
Answer: 0.3
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
No. The elementary formula divides by the probability of the given event and requires it to be positive.
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
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