Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn

Problem 2

What is the probability of getting (i) 2 white balls, (ii) 2 green balls, (iii) one of each colour in draw of two balls from a box containing 3 white and 4 green balls?

Solution

Total number of balls in the bag

= 3 White + 4 Green
= 7

Number of balls drawn = 2

Experiment : Drawing 2 balls from the bag

Total Number of Possible Choices

= Number of ways in which the two balls can be drawn from the total 7
⇒ n = 7C2
=
7 × 6
2 × 1
= 21

Let

  • A be the event of getting 2 white balls
  • B be the event of getting 2 green balls
  • C be the event of getting one ball of each color/colour i.e. 1 white and 1 green ball

For Event A

[White] [Green] Total
Available 3 4 7
To Choose 2 0 2
Choices 3C24C07C2

Number of Favorable Choices

= The number of ways in which the 2 white balls can be
drawn from the total 3
= 3C2
=
3 × 2
2 × 1
= 3

Probability of getting 2 white balls

⇒ Probability of occurrence of Event A

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(A) =
mA
n
=
3
21
=
1
7

• Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices
⇒ mAc = n − mA
= 21 − 3
= 18

» in favor

Odds in Favor of getting 2 white balls

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 3 : 18
= 1 : 6

» against

Odds against getting 2 white balls

⇒ Odds against Event A

= Number ofUnfavorable Choices : Number of Favorable Choices
= mAc : mA
= 18 : 3
= 6 : 1

For Event B

[White] [Green] Total
Available 3 4 7
To Choose 0 2 2
Choices 3C04C27C2

Number of Favorable Choices

= The number of ways in which the 2 green balls can be
drawn from the total 4
= 4C2
=
4 × 3
2 × 1
= 6

Probability of getting 2 green balls

⇒ Probability of occurrence of Event B

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(B) =
mB
n
=
6
21
=
2
7

• Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices
⇒ mBc = n − mB
= 21 − 6
= 15

» in favor

Odds in Favor of getting 2 green balls

⇒ Odds in Favor of Event B

= Number of Favorable Choices : Number of Unfavorable Choices
= mB : mBc
= 6 : 15
= 2 : 5

» against

Odds against getting 2 green balls

⇒ Odds against Event B

= Number ofUnfavorable Choices : Number of Favorable Choices
= mBc : mB
= 15 : 6
= 5 : 2

For Event C

[White] [Green] Total
Available 3 4 7
To Choose 1 1 2
Choices 3C14C17C2

Number of Favorable Choices

= The number of ways in which 1 white ball and 1 green ball can be
drawn from the total 7
= (Number of ways in which 1 white ball can be drawn from the total 3)
× (Number of ways in which 1 green ball can be drawn from the total 4)
= 3C1 × 4C1
= 3 × 4
= 12

Probability of getting a ball of each color i.e. a white ball and a green ball

⇒ Probability of occurrence of Event C

=
Number of Favorable Choices for the Event
Total Number of Possible Choices for the Experiment
⇒ P(C) =
mC
n
=
12
21
=
4
7

• Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices
⇒ mCc = n − mC
= 21 − 12
= 9

» in favor

Odds in Favor of getting a ball of each color

⇒ Odds in Favor of Event C

= Number of Favorable Choices : Number of Unfavorable Choices
= mC : mCc
= 12 : 9
= 4 : 3

» against

Odds against getting a ball of each color

⇒ Odds against Event C

= Number ofUnfavorable Choices : Number of Favorable Choices
= mCc : mC
= 9 : 12
= 3 : 4

Author : The Edifier