Probability practice · Easy

Option C (4) is correct because the probability that $N-2, \sqrt{3N}, N+2$ are in geometric progression is $\frac{1}{12}$, which simplifies to $\frac{4}{48}$, thus k=4.

Option D (None of the above is correct) is correct because independent events are not necessarily mutually exclusive, and the sum of their probabilities does not have to be 1.

Option D ($\frac{34}{99}$) is correct because it calculates the probability of a number being divisible by 4 or 7 within the given range using the principle of inclusion-exclusion.

Option B ($\frac{1}{8}$) is correct because it correctly applies the formula for the probability of the intersection of two events given their individual probabilities and the probability of their union.

Option C ($\frac{{{n^{n/2}}}}{{{n^{{n^2}/2}}}}$) is correct because it represents the probability of a binary operation being commutative, calculated by dividing the number of commutative binary operations by the total number of binary operations.

Option A (1/2) is correct because it calculates the conditional probability of at least one face showing 6, given that three dice show different faces.

The correct answer is ${ }^{16} C_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}$ because it calculates the probability of getting exactly six successes in the binomial distribution.

Option C ($\frac{3}{7}$) is correct because it accurately calculates the probability of forming a seven-digit number divisible by 2 using the given digits.

Option A (0.5) is correct because it is calculated by using the formula for conditional probability $P(F/E) = \frac{{P(E \cap F)}}{{P(E)}}$ after finding $P(E \cap F)$.

Option A ($\frac{{1}}{{12}}$) is correct because it is derived by using the given probabilities and the properties of conditional probability to find $P(B \cap C)$.

Option A ($\frac{1}{2}$) is correct because the probability of getting an odd number of heads in n tosses is always 1/2.

The correct answer is None of these because the probability of A is either 1/2 or 1/3, neither of which is provided as an option.

Option D ($P\left( M \right) + P\left( N \right) - 2P\left( {M \cap N} \right)$) is correct because it accounts for the probability of M occurring without N and N occurring without M.

Option C ($1 / 2$) is correct because the number of four-digit numbers containing unity is half the total number of four-digit numbers that can be formed.

Option C (b/(w+b)) is correct because the probability of drawing a black ball in the second draw is the same as the probability of drawing a black ball in the first draw, due to the ball being replaced with additional balls of the same color.

Option A ($P(\frac{{C}}{{D}}) \geq P(C)$) is correct because since $C \subset D$, $P(\frac{{C}}{{D}}) = \frac{{P(C)}}{{P(D)}}$ and $P(D) \le 1$, thus $P(\frac{{C}}{{D}}) \geq P(C)$.

Option B ($4/11$) is correct because there are 4 vowels out of 11 letters in the word 'PROBABILITY'.

Option B ($3 p^2-2 p^3$) is correct because it represents the probability of at least two of the independent events A, B, C occurring, given that P(A)=P(B)=P(C)=p.

Option A (0.5) is correct because it represents the probability that the bulb will last between 150 and 160 days, given the probabilities of lasting at least 150 days and at most 160 days.

Option A ($\frac{1}{14}$) is correct because it correctly calculates the conditional probability P(A/B) where A is the sum of digits being 8 and B is the product of digits being zero.

Option A ($\frac{{7}}{{12}}$) is correct because it calculates the probability of at least one red ball being drawn by subtracting the probability of no red balls being drawn from 1.

Option A (9) is correct because the probability that $x^2 + y^2$ is a multiple of 5 can be calculated by considering the last digits of x and y.

Option A (21/64) is correct because it calculates the probability of A or B hitting the target, but C not hitting the target, using independent probabilities.

Option D is correct because the probability is calculated by summing the favorable outcomes (at least three different numbers) and dividing by the total possible outcomes.

Option C (19/100) is correct. The probability that a randomly chosen card has a digit 5 is calculated by counting the number of cards with the digit 5 and dividing by the total number of cards.

Option A ($\frac{1}{{30}}$) is correct because it is the value of C that makes the sum of probabilities equal to 1, satisfying the condition for a probability function.

Option D ($\frac{5}{4}$) is correct because it's the variance calculated using the formula npq for a binomial distribution.

Option D ($\frac{\left({ }^{2 n} C_n\right)^2}{{ }^{4 n} C_{2 n}}$) is correct because it represents the number of ways to form two equal groups with n boys and n girls each, divided by the total number of ways to form two equal groups.

Option D (4) is correct because the probability of getting different numbers is 5/9, and q-p = 9-5 = 4.

Option A (0.4) is correct because it calculates the probability of exactly one event occurring by subtracting the probability of the intersection from the probability of the union.