1) A lotery has 1,000,000 tickets and draws 100 tickets to award prizes. How many tickets would you expect to have to purchase before you would win a prize?
2) Lesley claims she can roll triples with three dice in fewer than 20 rolls.
a)what is the expected number of rolls that it would take to roll the first set of triples?
b)what is the probability that she will roll the triples in fewer than five rolls?
Thanks for your help!
1) To determine how many tickets you would need to purchase before winning a prize, we need to calculate the expected number of tickets you would need to buy.
In this lottery, there are 1,000,000 tickets and 100 prizes, which means the probability of winning a prize on any individual ticket is 1/10,000 (100/1,000,000).
The expected value can be calculated by taking the reciprocal of the probability. In this case, it would be 1 divided by 1/10,000, which is equal to 10,000.
So, on average, you would expect to have to purchase 10,000 tickets before winning a prize.
2) a) To find the expected number of rolls it would take to roll the first set of triples with three dice, we need to calculate the average number of rolls needed.
The probability of rolling any specific triple (e.g., all threes) with three dice is 1/(6^3) = 1/216. This means that the probability of NOT rolling a triple on any specific roll is 1 - 1/216 = 215/216.
Let X be the number of rolls required to achieve the first set of triples. On the first roll, the probability of rolling a triple is 1/216. If we do not roll a triple, we need to start all over again. So, the expected number of rolls is:
E(X) = 1 + (215/216) * (E(X))
Simplifying this equation, we find:
E(X) = 216
Therefore, the expected number of rolls needed to roll the first set of triples is 216.
b) To find the probability that Lesley will roll the triples in fewer than five rolls, we can use the complement rule.
The probability of rolling the triples in five or more rolls would be equal to the sum of the probabilities of rolling no triples in each of the first four rolls:
P(no triples in first roll) = 215/216
P(no triples in second roll) = (215/216) * (215/216)
P(no triples in third roll) = (215/216) * (215/216) * (215/216)
P(no triples in fourth roll) = (215/216) * (215/216) * (215/216) * (215/216)
So, the probability of rolling the triples in fewer than five rolls is:
P(rolling triples in fewer than five rolls) = 1 - P(no triples in first roll) * P(no triples in second roll) * P(no triples in third roll) * P(no triples in fourth roll)
Remember that each roll is independent of the previous ones.
By substituting the values, we can calculate the probability.
1. p=0.0001
q=0.9999
P(x)=0.9999/0.0001
1. p=0.0001
q=0.9999
P(x)=0.9999/0.0001
=9999