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

Problem 5

In a bag there are ten balls of which three are red. Find the probability that in a draw of two balls (i) both will be red (ii) there will be at least one red ball.

Solution

Total number of balls in the bag

= 3 Red + 7 Others
= 10

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 12
⇒ n = 10C2
=
10 × 9
2 × 1
= 45

Let

  • A be the event of both the balls drawn being red
  • B be the event that there will be at least one red ball

For Event A

[Red] [Others] Total
Available 3 7 10
To Choose 2 0 2
Choices 3C27C010C2

Number of Favorable Choices

= The number of ways in which the 2 red balls can be
drawn from the total 10
= 3C2
=
3 × 2
2 × 1
= 3

Probability of both the balls drawn being red

⇒ 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
45
=
1
15

• Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices
⇒ mAc = n − mA
= 45 − 3
= 42

» in favor

Odds in Favor of both the balls drawn being red

⇒ Odds in Favor of Event A

= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 3 : 42
= 1 : 14

» against

Odds against both the balls drawn being red

⇒ Odds against Event A

= Number ofUnfavorable Choices : Number of Favorable Choices
= mAc : mA
= 3 : 45
= 1 : 15

For Event B

Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.

Event B can be accomplished in two alternative ways.

  • B1 : Drawing 1 Red Ball and 1 Other Ball
  • B2 : Drawing 2 Red Balls
[Red] [Others] Total
Available 3 7 10
To Choose 1 1 2 B1
Choices 3C17C110C2
To Choose 2 0 2 B2
Choices 3C27C010C2

» "B1"

Number of Favorable Choices

= The number of ways in which a red ball and another ball
can be drawn from the total 10
mB1 = (Number of ways in which 1 red ball can be drawn from the total 3)
× (Number of ways in which 1 other ball can be drawn from the total 7)
= 3C1 × 7C1
= 3 × 7
= 21

» "B2"

Number of Favorable Choices

= The number of ways in which 2 red balls can be drawn
from the total 10
mB2 = mA
= 3

Total number of Favorable/Favorable choices for Event B

⇒ mB = mB1 + mB2
= 21 + 3
= 24

Probability of drawing the balls such that there will be at least one red ball

⇒ 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
=
24
45
=
8
15

• Odds

Number of Unfavorable Choices

= Total Number of possible choices − Number of Favorable choices
⇒ mBc = n − mB
= 66 − 47
= 19

» in favor

Odds in Favor of drawing the balls such that they are of different colors/colours

⇒ Odds in Favor of Event B

= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mBc
= 47 : 19

» against

Odds against drawing the balls such that they are of different colors/colours

⇒ Odds against Event A

= Number ofUnfavorable Choices : Number of Favorable Choices
= mBc : mB
= 19 : 47