pakaskwa wrote:
c) The probability to win the 2nd prize is the same as that of the 1st prize, 1/10
I agree with your solutions to a) and b).
For c), I interpret the question differently; when it says "*only* the second prize", I take that to mean "the second prize but *not* the first prize". If we take that interpretation, the other two tickets need to be losing tickets, and the probability should be (1/10)*(28/29)*(27/28) = 27/290.
Alternatively, for c), you could use the results from a) and b). The probability that he wins the second prize is 1/10 (the same as the probability he wins the first prize). Then,
Prob(he wins *only* the 2nd prize) = Prob(he wins 2nd prize) - Prob(he wins both prizes) = (Answer from question a) - (Answer from question b)
= 1/10 - 1/145 = 27/290.
pakaskwa wrote:
d) Since there're 2 prizes available in 30 tickets, probability to win a prize with 1 ticket is 2/30=1/15. With 3 tickets, his probability to win is 3*1/15=1/5
For d), you can apply a simple logical test to see that the method is not quite right. By the method above, if he bought 15 tickets, he would be absolutely certain to win a prize, and if he bought 18 tickets, he'd have a 120% chance of winning a prize. Since he won't be certain to win a prize unless he buys 29 tickets, the method above cannot be correct. We could answer the question as follows:
Prob(he wins at least one prize) = 1 - Prob(he wins no prizes) = 1 - Prob(all three tickets lose)
= 1 - (28/30)(27/29)(26/28)
= 1 - (9/10)(26/29)
= 28/145
Alternatively, we could use the results from b) and c) above. From c), his probability of *only* winning the first prize is 27/290; his probability of *only* winning the second prize is 27/290; and his probability of winning both prizes, from b), is 2/290. So his probability of winning a prize = Prob(wins only 1st prize) + Prob(wins only 2nd prize) + Prob(wins both prizes) = 27/290 + 27/290 + 2/290 = 56/290 = 28/145. I actually did that with a Venn diagram, but it's hard to reproduce here.
