Chapter Closure
Sign In
search
Exercise 147 Page 701
a The number of combinations _nC_r in which we choose r objects from n possible is _nC_r = n!r!(n-r)!.
b The number of combinations _nC_r in which we choose r objects from n possible is _nC_r = n!r!(n-r)!.
c The number of combinations _nC_r in which we choose r objects from n possible is _nC_r = n!r!(n-r)!.
a About 1.81 %.
b About 18.18 %.
c About 6.81 %
a We have a jar with 5 red, 4 white, and 3 blue balls. We want to find the probability of choosing 3 white balls from the jar. This probability is the ratio of the number of ways to choose 3 white balls from the jar and the number of ways to choose 3 balls from the jar. The order of choosing the balls is not important, so each selection is a combination.
|
Combination |
|
Selection of objects in which the order is not important. |
_n C_r = n!/r! (n-r)!
SubstituteII
n= 4, r= 3
_4 C_3 = 4!/3! *( 4- 3)!
SubTerm
Subtract term
_4 C_3 = 4!/3! * 1!
n!=n* (n-1)!
_4 C_3 = 4 * 3!/3! * 1!
ReduceFrac
a/b=.a /3!./.b /3!.
_4 C_3 = 4/1!
1!=1
_4 C_3 = 4/1
CalcQuot
Calculate quotient
_4 C_3 = 4
_n C_r = n!/r! (n-r)!
SubstituteII
n= 12, r= 3
_(12) C_3 = 12!/3! *( 12- 3)!
SubTerm
Subtract term
_(12) C_3 = 12!/3! * 9!
▼
Simplify right-hand side
n!=n* (n-1)!
_(12) C_3 = 12 * 11!/3! * 9!
n!=n* (n-1)!
_(12) C_3 = 12 * 11 * 10!/3! * 9!
n!=n* (n-1)!
_(12) C_3 = 12 * 11 * 10 * 9!/3 * 2! * 9!
ReduceFrac
a/b=.a /9!./.b /9!.
_(12) C_3 = 12 * 11 * 10/3 * 2!
2!=2
_(12) C_3 = 12 * 11 * 10/3 * 2
Multiply
Multiply
_(12) C_3 = 1320/6
CalcQuot
Calculate quotient
_(12) C_3 =220
b We want to choose 2 red balls and 1 white ball from the jar. From Part A, we know that each selection of balls is a combination, and that there are 220 ways of choosing 3 balls from the jar. First, let's find the number of ways of choosing 2 red balls from 5 possible. Each such selection is a combination.
|
Combination |
|
Selection of objects in which the order is not important. |
_n C_r = n!/r! (n-r)!
SubstituteII
n= 5, r= 2
_5 C_2 = 5!/2! *( 5- 2)!
SubTerm
Subtract term
_5 C_2 = 5!/2! * 3!
n!=n* (n-1)!
_5 C_2 = 5 * 4!/2! * 3!
n!=n* (n-1)!
_5 C_2 = 5 * 4 * 3!/2! * 3!
ReduceFrac
a/b=.a /3!./.b /3!.
_5 C_2 = 5 * 4/2!
2!=2
_5 C_2 = 5 * 4/2
Multiply
Multiply
_5 C_2 = 20/2
CalcQuot
Calculate quotient
_5 C_2 = 10
c We want to find the probability of choosing 3 balls of the same color from the jar. This probability is the ratio between the number of ways of choosing 3 balls of the same color and the number of ways of choosing 3 balls. Note that each such selection is a combination.
|
Combination |
|
Selection of objects in which the order is not important. |
Red Balls
Since there are 5 red balls, choosing 3 red balls from the jar is the same as choosing which 2 balls should be left within the jar. In Part B, we found that there are 10 ways of choosing 2 red balls. Therefore, there are also 10 ways of choosing 5-2 = 3 red balls.
Blue Balls
There are three blue balls, so there is only 1 way of choosing three blue balls — we have to choose all of them.
Probability
We have 4 ways of choosing three white balls, 10 ways of choosing three red balls, and 1 ways of choosing three blue balls. Let's find the sum of this numbers. 4+ 10+ 1 = 15 In Part A, we found that there are 220 ways of choosing three balls from the jar. Let's divide the number of ways of choosing three balls of the same color ( 15) by the number of ways of choosing three balls from the jar ( 220), to get the desired probability. We will write the above probability as a percent by multiplying its decimal value by 100 %. P = 0.068181... * 100 % ≈ 6.81 %Loading content