Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705
Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.
Answer:
2x³ + x² - 11x - 15
Step-by-step explanation:
Step 1: Write out expression
2x³ + x² + 7x - 6 - (-2x + 10x + 10x + 9)
Step 2: Distribute negative
2x³ + x² + 7x - 6 + 2x - 10x - 10x - 9
Step 3: Combine like terms (x)
2x³ + x² + 9x - 10x - 10x - 9 - 6
2x³ + x² - x - 10x - 9 - 6
2x³ + x² - 11x - 9 - 6
Step 4: Combine like terms (constants)
2x³ + x² - 11x - 15
Answer:
Triangle ️ wdym if im doing it wrong
Equation: 50+15x=200
solution: x= 10
it would take him 10 weeks to purchase the bike he wants.
Answer:
Example:
A bag contains 3 black balls and 5 white balls. Paul picks a ball at random from the bag and replaces it back in the bag. He mixes the balls in the bag and then picks another ball at random from the bag.
a) Construct a probability tree of the problem.
b) Calculate the probability that Paul picks:
i) two black balls
ii) a black ball in his second draw
Solution:
tree diagram
a) Check that the probabilities in the last column add up to 1.
b) i) To find the probability of getting two black balls, first locate the B branch and then follow the second B branch. Since these are independent events we can multiply the probability of each branch.
ii) There are two outcomes where the second ball can be black.
Either (B, B) or (W, B)
From the probability tree diagram, we get:
P(second ball black)
= P(B, B) or P(W, B)
= P(B, B) + P(W, B)
