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Probability

Probability

MODULES

Basics of Probability
Mutually Exclusive Events
Non-Mutually Exclusive Events
Independent Events
Dependent Events
Conditional Probability
Odds For & Against
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Probability : Level 1
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ALL MODULES

CAT 2025 Lesson : Probability - Odds For & Against

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

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

Odds against event E
=n(E‾)n(E)=P(E‾)P(E)= \dfrac{n(\overline{\text{E}})}{n(\text{E})} = \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})}=n(E)n(E)​=P(E)P(E)​

For instance, if the probability of India winning a match is
45\dfrac{4}{5}54​, then for every 444 wins, there should be 111 loss.

∴\therefore∴ Odd for India winning is 4\bm{4}4 to 1\bm{1}1, while Odds against India winning is 1\bm{1}1 to 4\bm{4}4.

Example 21

The odds in favour of OnePlus launching a new phone is 444 to 333, while the odds against Apple launching a new phone is 222 to 555. 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 47\dfrac{4}{7}74​ and 57\dfrac{5}{7}75​ respectively.

Probability of neither of them launching
=(1−47)×(1−57)=37×27=649= \left(1 - \dfrac{4}{7}\right) \times \left(1 - \dfrac{5}{7}\right) = \dfrac{3}{7} \times \dfrac{2}{7} = \dfrac{6}{49}=(1−74​)×(1−75​)=73​×72​=496​

Probability of at least one of them launching
=1−649=4349= 1 - \dfrac{6}{49} = \dfrac{43}{49}=1−496​=4943​

Odds in favour of at least one of them launching
=43= 43=43 to (49−43)=43(49 - 43) = \bm{43}(49−43)=43 to 6\bm{6}6

Answer:
434343 to 666

Example 22

The odds for CSK defeating MI is 444 to 111. CSK and MI play 333 friendly games. What are the odds against CSK defeating MI in all 333 games?

(
111) 616161 to 646464            (222) 646464 to 616161           (333) 646464 to 125125125           (444) 125125125 to 646464          

Solution

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

P(CSK beating MI) =44+1=45= \dfrac{4}{4 + 1} = \dfrac{4}{5}=4+14​=54​

P(CSK beating MI in all
333 games) =45×45×45=64125= \dfrac{4}{5} \times \dfrac{4}{5} \times \dfrac{4}{5} = \dfrac{6 4}{125}=54​×54​×54​=12564​-----(1)

P(CSK not beating MI in all
333 games) =1−64125=61125= 1 - \dfrac{64}{125} = \dfrac{61}{125}=1−12564​=12561​ -----(2)

Odds against CSK beating MI in all
333 games =P(E‾)P(E)=(2)(1)=6164= \dfrac{\text{P}(\overline{\text{E}})}{\text{P}(\text{E})} = \dfrac{(2)}{(1)} = \dfrac{61}{64}=P(E)P(E)​=(1)(2)​=6461​

Answer: (
111) 616161 to 646464

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