In this topic you update probabilities after a fact is known. You keep only the outcomes that match the fact. You divide by the probability of the fact. You also multiply stage probabilities, average over cases, and reverse a conditional with Bayes' rule.
A problem belongs here when it says "given that". It also belongs here when it gives probabilities from cause to result and asks about the cause after seeing the result.
You will practice: Conditional probability, Multiplication rule, Law of total probability, Bayes' rule.
12 problems, each with a worked answer. Verified by an independent second solve.
Work them on paper, in order; difficulty ramps gently.
1.
Basic
Two fair six-sided dice are rolled. Given that the sum of the two dice is 8, what is the probability that at least one of the dice shows a 6?
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Answer: 52
The reduced sample space where the sum is 8 has 5 equally likely outcomes: (2,6), (3,5), (4,4), (5,3), (6,2). Exactly 2 of them, (2,6) and (6,2), show a 6.
P(at least one 6∣sum=8)=52
2.
Basic
Maya cycles to campus on dry days and takes the bus when it rains. On any given day it rains with probability 0.3. When it rains she is late with probability 0.2; when it does not rain she is late with probability 0.05. What is the probability that Maya is late on a randomly chosen day?
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Answer: 0.095
Partition on the weather and apply the law of total probability.
P(late)=0.3×0.2+0.7×0.05=0.06+0.035=0.095
3.
Basic
One card is drawn at random from a standard 52-card deck. Given that the card drawn is a face card (jack, queen, or king), what is the probability that it is a queen?
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Answer: 31
Conditioning on a face card shrinks the sample space to 12 equally likely cards. Of these, 4 are queens.
P(queen∣face card)=124=31
4.
Basic
A playlist contains 4 rock songs and 6 jazz songs, played in a uniformly random order. What is the probability that the first two songs played are both rock songs?
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Answer: 152
Multiply along the stages. Given the first song is rock, one rock song is used up for the second draw.
P=104⋅93=9012=152
5.
Basic
A fair coin is tossed three times. Given that at least two of the tosses came up heads, what is the probability that all three tosses came up heads?
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Answer: 41
Of the 8 equally likely outcomes, exactly 4 have at least two heads (HHH, HHT, HTH, THH) and 1 has three heads. Apply the definition of conditional probability.
P(HHH∣at least two heads)=4/81/8=41
6.
Basic
In a class of 30 students, 18 study Spanish, 12 study German, and 6 study both languages. A student is selected uniformly at random. Given that the selected student studies Spanish, what is the probability that they also study German?
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Answer: 31
Divide the probability of studying both languages by the probability of studying Spanish.
P(German∣Spanish)=18/306/30=186=31
7.
Basic
At a bakery, 60% of customers buy coffee. 30% of customers buy both coffee and a croissant. A customer is chosen at random. Given that this customer buys coffee, what is the probability that they also buy a croissant?
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Answer: 21
Apply the definition of conditional probability.
0.600.30=21
8.
Basic
A gardener plants one seed. The seed sprouts with probability 0.8. Given that it sprouts, it produces a flower with probability 0.5. What is the probability that the seed sprouts and produces a flower?
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Answer: 0.4
Multiply the sprout probability by the conditional flower probability.
0.8×0.5=0.4
9.
Basic
A ferry carries 40 vehicles: 25 cars and 15 vans. Exactly 10 of the cars and 6 of the vans are electric. One vehicle is selected uniformly at random. Given that the selected vehicle is electric, what is the probability that it is a van?
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Answer: 83
There are 10 + 6 = 16 electric vehicles. Restrict the sample space to them.
166=83
10.
Basic
A jar holds 3 green gumballs and 5 yellow gumballs. Two gumballs are drawn one after the other, without replacement, each uniformly at random. What is the probability that the first is yellow and the second is green?
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Answer: 5615
Use the multiplication rule for the two draws without replacement.
85×73=5615
11.
Basic
A print shop sends 60% of jobs to a laser printer and 40% to an inkjet printer. The laser printer misprints a job with probability 0.05. The inkjet printer misprints a job with probability 0.10. What is the probability that a randomly chosen job is misprinted?
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Answer: 0.07
Apply the law of total probability over the two printers.
0.6×0.05+0.4×0.10=0.03+0.04=0.07
12.
Basic
A beekeeper sells honey from two apiaries. 70% of jars come from the hillside apiary and 30% from the valley apiary. A hillside jar crystallizes with probability 0.2. A valley jar crystallizes with probability 0.6. A randomly chosen jar has crystallized. What is the probability that it came from the valley apiary?
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Answer: 169
Use Bayes' rule with the law of total probability in the denominator.
0.7×0.2+0.3×0.60.3×0.6=0.320.18=169