Question

Assume there is a fair lottery with 100 tickets sold.
The probability of any particular ticket winning is 0.01.
The probability that "Ticket number 1 will lose" is .99
So, what is the probability that "Ticket number 1 will lose" and "Ticket number 2 will lose"?

Answer 1

Identify the events as
A= ticket 1 looses
B= ticket 2 loosses

P(A)=0.99, P(b)=0.99
these are independent events so Probability that they both loose is P(A)*P(B)

Answer 2

Thank you ! I appreciate it.

Answer 3

please verify the answer or method , make sure we are not missing any information

Answer 4

Ah, I'm not sure what are you asking for. I honestly don't know the answer or the method (and nobody is grading this problem). That said, I have a weird problem then.

Consider if I was asking about 99 of those 100 tickets losing (instead of 2 of those 100), the method you gave would be .99 ^ 99, which results in 0.369. But, saying that exactly 99 particular tickets will lose is equivalent to saying the last 1 ticket will win. The chance of 1 ticket winning is .01, not .369.

Answer 5

It would make more sense to say P(A)=99/100, P(b)=98/99. So that P(A)*P(B) = .98

Right?

Answer 6

valid point but then they all cant have same probability of loosing or winning