Probability practice · Easy

Option B ($\frac{2}{9}$) is correct because it represents the conditional probability that the first ball is red, given that the second ball is red.

Option A (25/216) is correct because it represents the probability of not getting a six in the first two tosses and then getting a six on the third toss.

Option C ($\frac{15}{56}$) is correct because it is the probability of drawing exactly one red ball out of three, calculated using combinations.

Option B (0.2) is correct because it calculates the probability of a student failing both exams by subtracting the probability of passing at least one exam from 1.

Option B ($\frac{1}{5}$) is correct because it calculates the probability of a number chosen from the first 120 natural numbers being a multiple of 5 or 15.

Option A ($\frac{5}{12}$) is correct because it accurately calculates the probability of two randomly drawn diagonals intersecting inside the hexagon.

Option C (49/4) is correct because the expected value of the product is the product of the expected values of each throw, given the independence of the throws.

Option A (Only I is false) is correct because the probability of drawing an ace of spades is 1/52, not 2/52, while the probability of drawing a black card is indeed 26/52.

Option C ($\frac{3}{7}$) is correct because it calculates the probability of a leap year containing either 53 Sundays or 53 Mondays by considering the possible combinations of the two extra days in a leap year.

Option D (1/75) is correct because the probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

Option A (1/5) is correct because there are 4 favorable outcomes (a+b=9) out of a total of 20 possible outcomes in the Cartesian product A x B.

Option C ($\frac{{39}}{{50}}$) is correct because it is the calculated probability that the missing card is not a spade, given that two drawn cards are spades.

Option D ($\frac{5}{38}$) is correct because it represents the probability of drawing an I, then another I, and then a T in that specific order.

Option A is correct because $P(A \cap B)$ is always greater than or equal to $P(A) + P(B) - 1$ due to the properties of probability.

Option D (15/16) is correct because it calculates the probability of at least one head by subtracting the probability of all tails from 1.

Option B (Also independent) is correct because if A and B are independent, then A and the complement of B are also independent.

Option C ($3/28$) is correct because it represents the probability of five boys and three girls sitting in a row with no boy between two girls, calculated by considering the girls as one unit.

Option C (2/5) is correct because it calculates the probability of A and E being together by considering them as a single unit.

Option B ($\frac{53}{76}$) is correct because it calculates the probability of the product being divisible by 4 by finding the complement (not divisible by 4) and subtracting from 1.

Option A (2/7) is correct because it correctly uses the independence of events A and B to find P(B) from the given probabilities.

Option A ($\frac{3}{8}$) is correct because it is calculated using Bayes' theorem, considering the probability of the man speaking the truth and the probability of him lying.

Option B ($\left(\frac{{4}}{{5}}\right)^4$) is correct because it represents the probability of exactly 4 swimmers out of 5 students, accounting for the combination and probability of success.

Option D ($\frac{{15}}{{38}}$) is correct because it is calculated by finding the number of ways to choose 2 'I's and 1 'T' and dividing it by the total number of ways to choose 3 cards.

Option B ($\frac{83}{150}$) is correct because it calculates the probability that the product of two randomly selected numbers is divisible by 3 by considering the complement event (neither number is divisible by 3).

Option D ($\frac{5}{{12}}$) is correct because it calculates the probability of the first die showing a smaller number than the second die.

The correct answer is $1 / 2$ because it calculates the probability of getting at least two successes when tossing a die thrice.

Option B ($\frac{{ }^n P_m}{n^m}$) is correct because it represents the probability that no two passengers alight at the same floor when m passengers start in an elevator and stop at n floors.

Option D (90/216) is correct because it represents the probability of exactly two dice showing the same number when three dice are tossed.

Option A ($\frac{1}{19}$) is correct because after placing the first ball, there are 19 remaining spots, and only 4 of them are in the same box, so the probability is 4/19. However, the solution provided calculates the probability of any two balls being in the same box, not two specific balls. The correct probability is 1/19.

Option D ($\frac{{25}}{{36}}$) is correct because it is the conditional probability of needing at least 6 tosses given that more than 2 tosses are already needed.