Answer to Question #129647 in Statistics and Probability for Nahashon Kimathi

Question

A box contains 2 bags; bag X and bag Y. Bag X contains 2 red and 1 white beads. Bag Y contains 2 red and 2 white beads. A bag is selected at random. 2 beads are selected, one after the other without replacement. Determine the probability that:
A.2 red beads are selected.
B.1 red and 1 white bead are selected.
C.At least 1 red bead.

Answer 1

Solution:

Let D_XX be the event that the bag X is selected twice.

Let D_YY be the event that the bag Y is selected twice.

Let D_XY be the event that the bag X is selected once and the bag Y is selected once.

There are 4 options to choose from two bags twice: XX,XY,YX,YY

So,

P(D_XX)=1/4

P(D_XY)=1/2

P(D_YY)=1/4

A. Let A be event that 2 red beads are selected.

Then

P(A)=P(A|D_XX)P(D_XX)+P(A|D_XY)P(D_XY)+P(A|D_YY)P(D_YY)

P(A|D_XX)=m/n

m=C²₂=2!/2!/0!=1

n=C²₃=3!/2!/1!=3

P(A|D_XX)=1/3

P(A|D_XY)=m/n

m=C¹₂*C¹₂=2!/1!/1!*2!/1!/1!=2*2=4

n=(C¹₃)(C¹₄)=3!/1!/2!*4!/1!/3!=3*4=12

P(A|D_XY)=4/12=1/3

P(A|D_YY)=m/n

m=C²₂=1

n=C²₄=4!/2!/2!=6

P(A|D_YY)=1/6

P(A)=(1/3)*(1/4)+(1/3)*(1/2)+(1/6)*(1/4)

P(A)=7/24

The probability that 2 red beads are selected is equal 7/24

B. Let B be event that 1 red and 1 white bead are selected.

Then

P(B)=P(B|D_XX)P(D_XX)+P(B|D_XY)P(D_XY)+P(B|D_YY)P(D_YY)

P(B|D_XX)=m/n

m=C¹₂C¹₁=2*1=2

n=C²₃=3

P(B|D_XX)=2/3

P(B|D_XY)=m/n

m=C¹₂*C¹₂ + C¹₁*C¹₂=2*2+1*2=6

n=(C¹₃)(C¹₄)=3*4=12

P(B|D_XY)=6/12=1/2

P(B|D_YY)=m/n

m=C¹₂ C¹₂=2*2=4

n=C²₄=4!/2!/2!=6

P(B|D_YY)=4/6=2/3

P(B)=(2/3)*(1/4)+(1/2)*(1/2)+(2/3)*(1/4)=7/12

The probability that 1 red and 1 white bead are selected is equal 7/12.

C. Let C be event that at least 1 red bead.

Then C=A+B, since A$\bigcap$B=$\varnothing$

So

P(C)=P(A)+P(B)

P(C)=7/24+7/12=21/24=7/8

P(C)=7/8

The probability that at least 1 red bead is equal 7/8.

Here C^m_n=n!/m!/(n-m)! - binomial coefficient.