# Thread: cards, and balls in bags questions

1. ## cards, and balls in bags questions

1. elvalute 3(5 choose 2)+3!

2. five red cards are number 1 to 5 and seven blue cards are numbered 1 to 7
(a)in how many diffrent ways can be 2 cards be selected
(b) in how many ways can 2 cards be seclected if only red cards are picked
two cards picked at random
(c) what is the posibility both cards are red
what is the probility both are number 5

3. a bag contains eight balls number 1 to 8 3 balls are picked and placed in a row
(a)how many 3 digit numbers can be formed
(b) how many numbers are odd and lie between 200 and 300
(c)what is the porbabitlity that the number is less then 300
(D)that the number formed in all even.

2. Originally Posted by dooozer
1. elvalute 3(5 choose 2)+3!

2. five red cards are number 1 to 5 and seven blue cards are numbered 1 to 7
(a)in how many diffrent ways can be 2 cards be selected
(b) in how many ways can 2 cards be seclected if only red cards are picked
two cards picked at random
(c) what is the posibility both cards are red
what is the probility both are number 5

3. a bag contains eight balls number 1 to 8 3 balls are picked and placed in a row
(a)how many 3 digit numbers can be formed
(b) how many numbers are odd and lie between 200 and 300
(c)what is the porbabitlity that the number is less then 300
(D)that the number formed in all even.
1. 3(5C2) + 3! = 36

2.
a) total cards =12 ---> 12C2=66
b) 5C2=10
c) possibility(both cards are red) = 10/66
probability(both are number 5) = 1/66

3.
a) 8P3 or 8C3*3!=336
b) 1,3,5,7 must be the last digit so the result must be odd,
example 211,213,215,217,221,....etc, so there are four digits possible in units. and eight possibilities in tens. hence 4*8=32
c) Digit in hundreds must be less than 3 i.e 1,2
if first digit is 1 last two digits possibilities are 8P2=56
hence the total numbers formed are 56*2=112
prob. = 112/336 = 1/3
d) last digit must be in 2,4,6,8
first two digits possibility= 8P2=56
so the result is 56*4=224