Answer:
Step-by-step explanation:
A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.
(a) How many different samples of size 4 pens are possible?
12C4=12!/(4!*8!)=495
(b) How many samples have 3 red pens and 1 black pen?
5C3*7C1
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
=>5C3*7C1=10*7=70
(c) How many samples of size 4 contain at least two red pens?
(5C2*7C2)+(5C3*7C1)+(5C4*7C0)
5C2=5!/(2!*3!)=10
7C2=7!/(2!*5!)=21
5C3=5!/(3!*2!)=10
7C1=7!/(1!*6!)=7
5C4=5!/(4!*1!)=5
7C0=7!/(0!*7!)=1
=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285
(d) How many samples of size 4 contain at most one black pen?
(7C1*5C3)+(7C0*5C4)
7C1=7!/(1!*6!)=7
7C0=7!/(0!*7!)=1
5C3=5!/(3!*2!)=10
5C4=5!/(4!*1!)=5
=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75
