Question 478952
1) In this problem you have correctly stated that there are 5 odd-numbered tickets (1, 3, 5, 7, and 9) and we are assuming that there is no replacement after the first ticket is drawn, so...
The probability of drawing an odd-numbered ticket on the first draw is:
{{{5/9}}} but now you have only 4 odd-numbered tickets left out of a new total of 8 tickets, so the probability of drawing an odd-numbered ticket on the second draw is:
{{{4/8}}} These two events (A and B) are dependent events and the total probability is:
{{{P(A and B) = P(A)*P(B)}}} so the probability of drawing two odd-numbered tickets (no replacement) is...
{{{(5/9)*(4/8) = 20/72}}} = {{{5/18 = 0.28}}}
2) In this problem, the events are independent since the cards are replaced after each drawing.
The probability of drawing three spades (there are 13 of these in a standard deck) from a deck of 52 cards is:
{{{(13/52)*(13/52)*(13/52) = (13/52)^3}}} = {{{0.015625}}}