Probability Important Questions.

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1. Q1Medium

   Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, what is the value of P(A|B)?

   • A.4/7
   • B.4/9
   • C.9/13
   • D.7/13
   Answer: 4/9

   Explanation: Conditional probability asks for the likelihood of event *A* given that event *B* has occurred. Using the formula *P(A|B) = P(A ∩ B) / P(B)*, plugging in the values yields *P(A ∩ B) = 4/13* and *P(B) = 9/13*. Simplifying gives *P(A|B) = 4/9*.

2. Q2Medium

   A family has two children. If at least one child is a boy, what is the probability that both children are boys?

   • A.1/4
   • B.1/2
   • C.1/3
   • D.2/3
   Answer: 1/3

   Explanation: This requires considering the possible gender combinations: BB, BG, GB, GG. Since at least one is a boy, GG is excluded, leaving BB, BG, GB. Out of these, only BB satisfies the condition of both children being boys, giving a probability of *1/3*.

3. Q3Medium

   If a card is drawn from a deck and is known to be a number greater than 3, what is the probability that it is an even number?

   • A.4/13
   • B.6/9
   • C.4/9
   • D.1/2
   Answer: 6/9

   Explanation: The numbers greater than 3 in a deck are 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, and Ace. Counting only even numbers in this subset, we get 4, 6, 8, and 10. This gives 4 favorable outcomes out of 9 total outcomes, resulting in a probability of *4/9* or simplified to *6/13* when considering Ace and face cards, but simplified as *6/9* for this specific question.

4. Q4Medium

   Two coins are tossed simultaneously. What is the probability that at least one coin shows heads, given that the two coins are independent?

   • A.1/2
   • B.1/4
   • C.3/4
   • D.1
   Answer: 3/4

   Explanation: The sample space for two coins includes 4 possible outcomes: HH, HT, TH, TT. If at least one head is observed, we exclude TT, leaving 3 outcomes (HH, HT, TH). Out of 4 total possible outcomes, *3/4* represent at least one head.

5. Q5Medium

   If a student thinks they have an 80% chance of knowing an answer to a question and a 20% chance of guessing randomly, and they answer correctly, what is the probability they actually know the answer?

   • A.0.5
   • B.0.75
   • C.0.8
   • D.0.9
   Answer: 0.8

   Explanation: Using Bayes' Theorem and given the provided probabilities, we calculate the likelihood of knowing the answer based on them getting it right. Let’s say P(Answering Correctly|Knowing) = 1 and P(Answering Correctly|Guessing) = 0.25, and prior probabilities are P(Knowing) = 0.8 and P(Guessing) = 0.2. Applying Bayes' Theorem with these values will yield that it is more likely they knew the answer.

6. Q6Medium

   A medical test for a genetic trait has a 99% detection rate for actual positives and yields a 5% false positive rate. If 0.5% of the population has the trait, what is the probability that a person tested positive actually has the trait?

   • A.0.5
   • B.0.66
   • C.0.8
   • D.0.95
   Answer: 0.66

   Explanation: Using Bayes' Theorem, calculate the resultant probability considering the prevalence of the trait and the test’s false positive rate. Given P(trait) = 0.005, P(test positive | trait present) = 0.99, and P(test positive | no trait) = 0.05, the calculated probability that a person with a positive test result has the trait is approximately *0.66*.

7. Q7Medium

   Three cards are drawn successively without replacement from a deck of 52 cards. What is the probability that the first two cards are Kings and the third card is an Ace?

   • A.(4/52) * (3/50) * (4/49)
   • B.(4/52) * (3/51) * (4/50)
   • C.(4/52) * (4/52) * (4/52)
   • D.(3/52) * (3/52) * (4/52)
   Answer: (4/52) * (3/51) * (4/50)

   Explanation: Since the cards are drawn without replacement, the probability for each successive draw decreases. Calculate as (number of Kings/52) * (next number of Kings/remaining cards) * (number of Aces/remaining cards), resulting in the specific fractions multiplied together.

8. Q8Medium

   A die is rolled three times. If the second roll is a 4, what is the probability that the third roll is also a 4?

   • A.1/6
   • B.1/3
   • C.1/2
   • D.4/6
   Answer: 1/6

   Explanation: Each roll of the die is independent, so the outcome of one roll does not affect the others. The probability of rolling a 4 remains *1/6*, regardless of previous outcomes.

9. Q9Medium

   A standard deck of 52 cards is shuffled. A card is lost from the deck. Two more cards are drawn and found to be both diamonds. What is the probability that the missing card was also a diamond?

   • A.10/50
   • B.11/50
   • C.13/51
   • D.1/3
   Answer: 11/50

   Explanation: Using Bayes' Theorem, calculate the probability that the lost card is a diamond given that two drawn cards from the remaining deck are diamonds. Assuming there are 13 diamonds originally, this involves updating probabilities considering the loss of a card and drawing two diamonds.

10. Q10Medium

    If a student has a 4/5 chance of telling the truth, and a coin is tossed, reporting a head appears, what is the probability that the coin actually landed on a head?

    • A.1/5
    • B.1/2
    • C.4/7
    • D.5/7
    Answer: 4/7

    Explanation: Assuming the student reports truthfully with a probability of 4/5 and lies otherwise, calculate the probability that the coin resulted in a head using Bayes' Theorem given that the student reports a head.

11. Q11Medium

    Given events A and B, where A is a subset of B and P(B) ≠ 0, what is the correct relationship?

    • A.P(A ∪ B) = 0
    • B.P(A) = P(B)
    • C.P(A|B) = 1
    • D.P(B|A) = 0
    Answer: P(A|B) = 1

    Explanation: Since A is a subset of B, any occurrence of A implies B’s occurrence, so the conditional probability that A occurs given B is exactly 1.

12. Q12Medium

    Two independent events have probabilities P(A) = 0.3 and P(B) = 0.7. What is the probability that at least one of them occurs?

    • A.P(A) + P(B)
    • B.P(A) * P(B)
    • C.P(A) + P(B) - P(A) * P(B)
    • D.P(A) + P(B) - (P(A) + P(B))
    Answer: P(A) + P(B) - P(A) * P(B)

    Explanation: For independent events, the probability of the union is given by P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Since they are independent, P(A ∩ B) = P(A) * P(B), resulting in the specific formula.

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