i Employment data at a large company reveal that 55 of the

i) Employment data at a large company reveal that 55 % of the
workers are married, that 41 % are college graduates, and
that 1/3 of the college graduates are married.
What is the probability that a randomly chosen worker is:
a) neither married nor a college graduate?
b) married but not a college graduate?
c) married or a college graduate?

ii)  box contains 21 yellow, 28 green and 40 red jelly beans.
If 13 jelly beans are selected at random, what is the probability that:
a) 10 are yellow?  
b) 10 are yellow and 2 are green?  
c) At least one is yellow?

Solution

1)

P ( Not married + not graduate) = 1 - (0.55 +0.41 - 0.41/3)

= 0.176

P( Married but not college graduate) = 0.55 - (0.41/3)

= 0.4133

P ( Married or college graduate) =

= 0.55 + (0.41/3)

=0.6866

2)

total jelly beans = 21 + 28 + 40 = 89

P(10 yellow) = ²¹C₁₀ * ⁶⁸C₃ / ⁸⁹C₁₃

= 1.257 * 10^-5

P ( 10 yellow, 2 green, 1 red) = ²¹C₁₀ * ²⁸C₂ * ⁴⁰C₁ / ⁸⁹C₁₃

= 3.793 * 10^-6

P ( At least 1 yellow) = 1 - P ( No yellow)

= 1 - ⁶⁸C₁₃ / ⁸⁹C₁₃

= 0.9776

Hope this helps.