CAT 2025 Lesson : Probability - Odds For & Against

The odds in favour of OnePlus launching a new phone is $4$ to $3$, while the odds against Apple launching a new phone is $2$ to $5$. 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 $= 43$ to $(49 - 43) = \bm{43}$ to $\bm{6}$

Answer: $43$ to $6$

The odds for CSK defeating MI is $4$ to $1$. CSK and MI play $3$ friendly games. What are the odds against CSK defeating MI in all $3$ games?

($1$) $61$ to $64$            ($2$) $64$ to $61$           ($3$) $64$ to $125$           ($4$) $125$ to $64$          

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: ($1$) $61$ to $64$