A bag contains tickets numbered 1 to 30.Three tickets are drawn at random from the bag.Find the probability that the maximum number on the selected tickets exceeds 25?

1 Answer
Apr 11, 2018

#0.4335#

Explanation:

#"The complementary event is that the maximum is equal or"#
#"less than 25, so that the three tickets are all three among the"#
#"first 25. The odds for that are : "#

#(25/30)(24/29)(23/28) = 0.5665#

#"So the asked probability is :"#

#1 - 0.5665 = 0.4335#

#"Further explanation : "#

#P(A and B and C) = P(A) P(B|A) P(C|AB)#
#"On the first draw the odds that the first ticket has number less"#
#"or equal than 25 is (25/30). So P(A) = 25/30."#
#"When drawing the second ticket,"#
#"there are only 29 tickets left in the bag and 5 of them have a"#
#"number bigger than 25 if the first ticket has number <= 25, so"#
#"P(B|A) = 24/29."#
#"For the third draw, there are 28 tickets left. 23 of them are"#
#"<= 25, if the previous draws were also <= 25, hence (23/28)."#
#"So P(C|AB) = 23/28."#