Conditional Probability - Problem Solving
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A lot of difficult probability problems involve conditional probability. These can be tackled using tools like Bayes' Theorem, the principle of inclusion and exclusion, and the notion of independence.
Two standard dice with 6 sides are thrown and the faces are recorded. Given that the sum of the two faces equals to 10, what is the probability that the first throw equals to 5?
Let \(A\) denote the event for which the two faces sums to 10, and \(B\) the event for which the first throw equals 5.
Then the sample space for \(A\) is \( \{ (5,5),(4,6),(6,4) \} \).
The sample space for \(B\) is \( \{ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \} \).Therefore, \(A \cap B = \{(5,5)\} \) and \(P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{1/36}{3/36} = \frac13. \ _\square \)
A bag contains a number of coins, one of which is a two-headed coin and the rest are fair coins. A coin is selected at random and tossed. If the probability that the toss results in a head is \(\frac {7}{12} \), then how many fair coins are in the bag ?
Details and Assumptions:
A two-headed coin is a coin which has head on both sides; a fair coin means it has tail on one side and head on the other.
Horace turns up at school either late or on time. He is then either shouted at or not. The probability that he turns up late is \(0.4.\) If he turns up late, the probability that he is shouted at is \(0.7\). If he turns up on time, the probability that he is still shouted at for no particular reason is \(0.2\).
You hear Horace being shouted at. What is the probability that he was late?
