You roll a fair six-sided die 7 times. What is the probability of getting a) exactly two 4’s?
b) less than (and not equal to) two 4’s? c) two or more 4’s?
To find the probability of getting a particular outcome in repeated dice rolls, you need to understand the concept of probability and basic counting principles.
a) To find the probability of getting exactly two 4's in 7 rolls of a fair six-sided die, we can utilize the concept of binomial probability. The general formula to calculate the probability of obtaining a specific number of successes (k) in n trials is:
P(k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
n C k represents the number of combinations of choosing k items from a set of n items. It can be calculated using the formula: n! / (k! * (n - k)!)
p is the probability of success in a single trial (rolling a 4 in this case), which is 1/6 for a fair six-sided die.
Using this formula, for exactly two 4's in 7 rolls, the probability is:
P(2) = (7 C 2) * (1/6)^2 * (5/6)^(7 - 2)
Calculating this, we find:
P(2) = (7! / (2! * (7 - 2)!)) * (1/6)^2 * (5/6)^5
= (21) * (1/36) * (3125/7776)
≈ 0.1615 (rounded to four decimal places)
Therefore, the probability of rolling exactly two 4's in 7 rolls of a fair six-sided die is approximately 0.1615.
b) To find the probability of rolling less than (and not equal to) two 4's, we need to calculate the probabilities of getting zero 4's and one 4, and then add them together.
P(0) = (7 C 0) * (1/6)^0 * (5/6)^(7 - 0)
= (1) * (1) * (78125/279936)
≈ 0.2789 (rounded to four decimal places)
P(1) = (7 C 1) * (1/6)^1 * (5/6)^(7 - 1)
= (7) * (1/6) * (15625/46656)
≈ 0.3114 (rounded to four decimal places)
P(less than 2) = P(0) + P(1)
≈ 0.2789 + 0.3114
≈ 0.5903 (rounded to four decimal places)
Therefore, the probability of rolling less than (and not equal to) two 4's in 7 rolls of a fair six-sided die is approximately 0.5903.
c) To find the probability of rolling two or more 4's, we can subtract the probabilities of rolling zero 4's and one 4 from 1 (total probability).
P(2 or more) = 1 - (P(0) + P(1))
= 1 - (0.2789 + 0.3114)
= 1 - 0.5903
≈ 0.4097 (rounded to four decimal places)
Therefore, the probability of rolling two or more 4's in 7 rolls of a fair six-sided die is approximately 0.4097.