In this topic you compute probabilities like both, none, exactly one, and at least one. When events are independent, the probability that both happen is the product of the two probabilities. Knowing one happened does not change the probability of the other. The cue is the word independently in the problem statement. Links fail independently. Each chip is defective independently. If the problem does not say it, do not multiply. You will practice: the multiplication rule, complements and at least one, exactly k by disjoint patterns, conditioning on built events.
A data center has two network links that fail independently of each other. On any given day, link 1 fails with probability 0.2 and link 2 fails with probability 0.3. What is the probability that at least one of the two links fails on a given day?
Answer:
By independence, first find the probability that neither link fails. Then take the complement.
A factory produces microchips, and each chip is defective with probability 0.1, independently of all other chips. An inspector examines 3 chips. What is the probability that none of the 3 chips is defective?
Answer:
Each chip is good with probability 0.9 and the chips are independent, so multiply.
On a given day it rains in Hilltown with probability 0.3 and in Baytown with probability 0.4, and the two events are independent. What is the probability that it rains in both towns on that day?
Answer:
Independence licenses multiplying the two probabilities.
A wildlife camera trap is set up in snow leopard territory. On each night, a leopard passes the camera with probability 1/5, independently of other nights. The camera runs for 3 nights. What is the probability it records a leopard on at least one night?
Answer:
Go through the complement of at least one visit.
A string duo is booked for a wedding. The violinist arrives late with probability 0.1 and the cellist arrives late with probability 0.2, independently of each other. What is the probability that both musicians arrive on time?
Answer:
The violinist is on time with probability 0.9 and the cellist with probability 0.8. Multiply by independence.
A hiker carries two separate GPS units on a trek. Each unit independently has a dead battery with probability 0.05. What is the probability that at least one unit works?
Answer:
The complement of at least one working is both batteries dead, and independence gives its probability as a product.
A basketball player makes each free throw with probability 0.7, independently of her other throws. She takes two free throws. What is the probability that she makes both?
Answer:
The throws are independent, so multiply.
Leo takes a bus and then a train to school. The bus is on time with probability 0.9 and the train is on time with probability 0.8, independently of each other. What is the probability that both are on time?
Answer:
Independence lets us multiply the two probabilities.
Nora plays a claw machine 3 times. Each play wins a prize with probability 0.3, independently of the other plays. What is the probability that she wins at least one prize?
Answer:
Use the complement: she wins no prize on all 3 plays.
A florist orders roses and tulips from two suppliers that deliver independently. The roses arrive on time with probability 0.8 and the tulips arrive on time with probability 0.5. What is the probability that the roses arrive on time but the tulips do not?
Answer:
The tulips are late with probability 0.5. Independence also holds for complements, so multiply.
Priya and Tom each attempt the same sudoku puzzle, working independently. Priya solves it with probability 0.6 and Tom solves it with probability 0.5. What is the probability that exactly one of them solves it?
Answer:
Add the two disjoint patterns: only Priya solves it, or only Tom does.
Three friends each buy a ticket in three different raffles. Independently, the first friend wins with probability 0.1, the second with probability 0.2, and the third with probability 0.25. What is the probability that at least one of the friends wins?
Answer:
Take the complement: nobody wins. Multiply the three losing probabilities.
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