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Modern Maths

Probability

ALL MODULES

CAT 2025 Lesson : Probability - Odds For & Against

7. Odds for and Odds Against

The odds for an event is the ratio of the number of favourable cases to the number of unfavourable cases.

Odds for event E

= \dfrac{n(\text{E})}{n(\overline{\text{E}})} = \dfrac{\text{P}(\text{E})}{\text{P}(\overline{\text{E}})}

The odds against an event is the ratio of number of unfavourable cases to the number of favourable cases.

Odds against event E

= \dfrac{n(\overline{\text{E}})}{n(\text{E})} = \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})}

For instance, if the probability of India winning a match is

\dfrac{4}{5}

, then for every

4

wins, there should be

1

loss.

\therefore

Odd for India winning is

\bm{4}

\bm{1}

, while Odds against India winning is

\bm{1}

\bm{4}

Example 21

The odds in favour of OnePlus launching a new phone is

4

3

, while the odds against Apple launching a new phone is

2

5

. What is the odds in favour of at least one of them launching a new phone?

Solution

Probability of OnePlus and Apple launching new phones are

\dfrac{4}{7}

and

\dfrac{5}{7}

respectively.

Probability of neither of them launching

= \left(1 - \dfrac{4}{7}\right) \times \left(1 - \dfrac{5}{7}\right) = \dfrac{3}{7} \times \dfrac{2}{7} = \dfrac{6}{49}

Probability of at least one of them launching

= 1 - \dfrac{6}{49} = \dfrac{43}{49}

Odds in favour of at least one of them launching

= 43

(49 - 43) = \bm{43}

\bm{6}

Answer:

43

6

Example 22

The odds for CSK defeating MI is

4

1

. CSK and MI play

3

friendly games. What are the odds against CSK defeating MI in all

3

games?

(

1

)

61

64

(

2

)

64

61

(

3

)

64

125

(

4

)

125

64

Solution

We should first convert the odds to probability and then apply it.

P(CSK beating MI)

= \dfrac{4}{4 + 1} = \dfrac{4}{5}

P(CSK beating MI in all

3

games)

= \dfrac{4}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{6 4}{125}

-----(1)

P(CSK not beating MI in all

3

games)

= 1 - \dfrac{64}{125} = \dfrac{61}{125}

-----(2)

Odds against CSK beating MI in all

3

games

= \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})} = \dfrac{(2)}{(1)} = \dfrac{61}{64}

Answer: (

1

)

61

64

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Back

Quant

Modern Maths

Probability

ALL MODULES

CAT 2025 Lesson : Probability - Odds For & Against

7. Odds for and Odds Against

The odds for an event is the ratio of the number of favourable cases to the number of unfavourable cases.

Odds for event E

= \dfrac{n(\text{E})}{n(\overline{\text{E}})} = \dfrac{\text{P}(\text{E})}{\text{P}(\overline{\text{E}})}

The odds against an event is the ratio of number of unfavourable cases to the number of favourable cases.

Odds against event E

= \dfrac{n(\overline{\text{E}})}{n(\text{E})} = \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})}

For instance, if the probability of India winning a match is

\dfrac{4}{5}

, then for every

4

wins, there should be

1

loss.

\therefore

Odd for India winning is

\bm{4}

\bm{1}

, while Odds against India winning is

\bm{1}

\bm{4}

Example 21

The odds in favour of OnePlus launching a new phone is

4

3

, while the odds against Apple launching a new phone is

2

5

. What is the odds in favour of at least one of them launching a new phone?

Solution

Probability of OnePlus and Apple launching new phones are

\dfrac{4}{7}

and

\dfrac{5}{7}

respectively.

Probability of neither of them launching

= \left(1 - \dfrac{4}{7}\right) \times \left(1 - \dfrac{5}{7}\right) = \dfrac{3}{7} \times \dfrac{2}{7} = \dfrac{6}{49}

Probability of at least one of them launching

= 1 - \dfrac{6}{49} = \dfrac{43}{49}

Odds in favour of at least one of them launching

= 43

(49 - 43) = \bm{43}

\bm{6}

Answer:

43

6

Example 22

The odds for CSK defeating MI is

4

1

. CSK and MI play

3

friendly games. What are the odds against CSK defeating MI in all

3

games?

(

1

)

61

64

(

2

)

64

61

(

3

)

64

125

(

4

)

125

64

Solution

We should first convert the odds to probability and then apply it.

P(CSK beating MI)

= \dfrac{4}{4 + 1} = \dfrac{4}{5}

P(CSK beating MI in all

3

games)

= \dfrac{4}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{6 4}{125}

-----(1)

P(CSK not beating MI in all

3

games)

= 1 - \dfrac{64}{125} = \dfrac{61}{125}

-----(2)

Odds against CSK beating MI in all

3

games

= \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})} = \dfrac{(2)}{(1)} = \dfrac{61}{64}

Answer: (

1

)

61

64

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