2 A box contains 10 blue and 5 red markers Find the probabil

2. A box contains 10 blue and 5 red markers. Find the probability of obtaining at least two red markers if four markers are randomly selected without replacement. /Hint: Consider the compliment event.]

Solution

P0 = P(no red among four)
= (10/15)(9/14)(8/13)(7/12)

P1 = P(exactly 1 red among four)
= P(red is the 1st) + P(red is the 2nd) + P(red is the 3rd) + P(red is the 4th)
= (5/15)(10/14)(9/13)(8/12) + (10/15)(5/14)(9/13)(8/12) + (10/15)(9/14)(5/13)(8/12) + ...
...+ (10/15)(9/14)(8/13)(5/12)
= 4 * (5/15)(10/14)(9/13)(8/12)
therefore :

P0 + P1 = (10/15)(9/14)(8/13)(7/12) + 4 * (5/15)(10/14)(9/13)(8/12)
= (10*9*8*7 + 4*5*10*9*8)/(15*14*13*12)
= 10*9*8(7 + 20)/(15*14*13*12)
# 0,59341
and finally :

P(at least 2 on four) = 1 - (P0 + P1)
# 0.4066 = 40.66%

hope it\' ll help !!