1An urn contains eight balls 2 red 2 blue 2 orange 2 green T

1.An urn contains eight balls 2 red 2 blue 2 orange 2 green, The balls are separated at random into two set of four balls each,what is the probability that each set contains one ball of each color?

2.A bin contains 100 balls numbered consecutively from 1 to 100 two balls are chosen at random without placement, what is the probability that the total of the numbers on the chosen balls is even?

3.Apoker hand contains five cards, find the probability that a poker hand be :

a:\"four of a kind\"(contains four cards of equal face value)

b:\"full house\"(three cards of equal face value and two other equal face value)

c:\"three of kind\"(three card have the equal face values different from the three from eachother)

d:\"one pair\"(two card has equal face value, and three cards with face value different from two and from each other)

e:\"two pairs\"(two cards of equal facevalue, two other cards of equal face value different from the value of the first two, and one card with face value different from each of other four)

Solution

1. 8 balls are separated into two sets with 4 balls each.

let A be the event that each set contain each ball of one colour. i.e. out of the four balls one should be red,one should be blue one should be orange and the other should be green.

now we can select 4 balls from 8 balls in ⁸C₄ ways.[all possible cases]

from 2 red balls we can select one red ball with ²C₁ ways. similarly for blue,green and orange.[number of favourable cases]

so P[A]=(²C₁*²C₁*²C₁*²C₁)/⁸C₄=0.22857 [answer]

2. let B be the event that the total number on the balls is even.

now sum of two numbers are even means either both of them are even or both of them are odd.

the balls are drawn without replacement.

now from 1 to 100 there are 50 odd numbers (1,3,5,7,9,11,.....,99) and 50 even numbers(2,4,6,8,10,....,100)

so the first ball number can be odd in 50/100 ways. as they are drawn without replacement the second ball is chosen from remaining 99 balls.

so P[B]=P[first ball number is odd and the second ball number chosen from remaining 99 balls with 49 odd numbered balls is also odd]+P[first ball number is even and the second ball number chosen from remaining 99 balls with 49 even numbered balls is also even]

           =(50/100)*(49/99)+(50/100)*(49/99)=0.4949 [answer]

3. a poker hand contains 5 cards. so we can select five cards from 52 cards in ⁵²C₅ ways.

a) four cards of equal face values. that means these 4 are selected from the 4 cards of a particular face value from 13 types of face value.the other one is selected from the remaining 48 cards.

so P[four of a kind]=⁽⁴C₄*⁴⁸C₁)/⁵²C₅=0.000018 [answer]

b) three cards are of equal face value means these three are chosen from the 4 cards of a particluar face value from 13 types of face value in ⁴C₃ ways.the other two cards can be selected from any one of 12 remaining faces and can be chosen in ⁴C₂ ways.

P[full house]=(⁴C₃*⁴C₂*¹²C₁)/⁵²C₅=0.00011 [answer]

c) three cards are of equal face value means these three are chosen from the 4 cards of a particluar face value of 13 types in ⁴C₃ ways.the 4th one can be chosen from 48 cards and the last can be chosen from 44 cards as it has to be different from all four.these 5 cards.

so P[three of a kind]=(⁴C₃*⁴⁸C₁*⁴⁴C₁)/⁵²C₅=0.00325 [answer]

d) two cards have equal face values in ⁴C₂ ways (from similar logics as above). the third card can be anything of 48 remaining cards. the 4th one can be anything of the remaining 44 cards and the 5th one can be anything of the remaining 40 cards as they are all different.

so P[one pair]=(⁴C₂*⁴⁸C₁*⁴⁴C₁*⁴⁰C₁)/⁵²C₅=0.195 [answer]

e) two cards have equal face values in ⁴C₂ ways (from similar logics as above).the other two have different equal face values in 12* ⁴C₂ ways. the last one can be anything of the 44 cards.

P[two pairs]=⁴C₂*12* ⁴C2*⁴⁴C₁/⁵²C₅ =0.00731 [answer]