3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Three fair six-sided dice are rolled; the sample space is the set of 216 equally likely ordered triples. Given that the product of the three numbers shown is even, the probability that their sum is odd is ______. Exactly one option is correct.
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Answer: C
Let E = product even and O = sum odd. The complement of E is 'all three dice odd': 33=27 triples, so n(E)=216−27=189.
The sum is odd exactly when an odd number of the dice (one or three) show odd numbers. If all three are odd the product is odd, so E∩O = 'exactly one die odd': choose its position (3), its value (3), and the two even values (3⋅3): n(E∩O)=3⋅3⋅9=81.
P(O∣E)=18981=73.
Trap: 21681=83 is P(E∩O), not the conditional probability; 21 ignores the conditioning.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Box I contains 3 white and 2 black balls; box II contains 2 white and 3 black balls. A fair coin is tossed to choose one box, and from that box two balls are drawn one after the other without replacement. Both are white. A third ball is now drawn from the same box (from the three balls left in it). The probability that the third ball is white is ______. Exactly one option is correct.
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Answer: A
P(WW∣I)=53⋅42=103, P(WW∣II)=52⋅41=101.
Bayes: P(I∣WW)=21⋅103+21⋅10121⋅103=43, P(II∣WW)=41.
After WW, box I holds 1 white, 2 black; box II holds 0 white, 3 black. By total probability (conditioned on WW):
P(3rd W∣WW)=43⋅31+41⋅0=41.
(Check: P(WWW)=21⋅103⋅31=201, P(WW)=51, ratio 41.)
Trap: using the prior 21 for box I instead of the posterior gives 61.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Four married couples (eight distinct people) sit at random around a round table with eight seats, all 7! circular arrangements (arrangements that differ only by a rotation being identified) being equally likely. The probability that no husband sits next to his own wife is ______. Exactly one option is correct.
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Answer: D
Sample space: 7!=5040 circular arrangements, equally likely.
Let Ai = couple i sit together. Treating k specified couples as blocks (each block in 2 internal orders), there are 8−k units around the table: n(Ai1∩⋯∩Aik)=2k(8−k−1)!.
Inclusion-exclusion: n(no couple together)=∑k=04(−1)k(k4)2k(7−k)!
=5040−4⋅2⋅720+6⋅4⋅120−4⋅8⋅24+16⋅6=5040−5760+2880−768+96=1488.
Probability =50401488=10531. (Confirmed by enumerating all 5040 arrangements.)
Trap: forgetting the factor 2k for the internal order of each couple, or using (8−k)! (linear arrangements) for a round table.
Difficulty: Peak challenge