If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
6 ∶ 5
Calculating Odds Against an Event
The question asks us to determine the odds against a random event A, given that the odds in favour of event A are 5 ∶ 6.
Understanding Odds in Probability
In probability, odds represent the ratio of outcomes rather than the ratio of outcomes to the total possible outcomes (which is what probability does). There are two main types of odds:
- Odds in favour: This is the ratio of the number of favourable outcomes to the number of unfavourable outcomes.
- Odds against: This is the ratio of the number of unfavourable outcomes to the number of favourable outcomes.
Applying the Concepts to Event A
We are given that the odds in favour of event A are 5 ∶ 6.
This can be interpreted as:
- Number of favourable outcomes for event A = 5
- Number of unfavourable outcomes for event A = 6
The total number of possible outcomes is the sum of favourable and unfavourable outcomes:
\(\text{Total Outcomes} = \text{Favourable Outcomes} + \text{Unfavourable Outcomes}\)
\(\text{Total Outcomes} = 5 + 6 = 11\)
Determining the Odds Against Event A
The odds against event A are defined as the ratio of the number of unfavourable outcomes to the number of favourable outcomes.
\(\text{Odds Against A} = \frac{\text{Number of Unfavourable Outcomes}}{\text{Number of Favourable Outcomes}}\)
\(\text{Odds Against A} = \frac{6}{5}\)
Therefore, the odds against event A are 6 ∶ 5.
We can also relate odds to probability. If the odds in favour of A are 5 ∶ 6:
- Probability of A (P(A)) = \(\frac{\text{Favourable Outcomes}}{\text{Total Outcomes}} = \frac{5}{11}\)
- Probability of not A (P(A')) = \(\frac{\text{Unfavourable Outcomes}}{\text{Total Outcomes}} = \frac{6}{11}\)
The odds against A can also be expressed as the ratio of P(A') to P(A):
\(\text{Odds Against A} = \frac{P(A')}{P(A)} = \frac{6/11}{5/11} = \frac{6}{5}\)
Both methods confirm that the odds against event A are 6 ∶ 5.
| Concept | Ratio | Description |
|---|---|---|
| Odds in Favour of A | 5 ∶ 6 | Favourable ∶ Unfavourable |
| Odds Against A | 6 ∶ 5 | Unfavourable ∶ Favourable |
Revision Table: Odds Terminology
This table summarises the key terms related to odds.
| Term | Formula (using F=Favourable, U=Unfavourable) | Example (Odds in Favour 5:6) |
|---|---|---|
| Odds in Favour | F ∶ U | 5 ∶ 6 |
| Odds Against | U ∶ F | 6 ∶ 5 |
| Probability of Event | F / (F + U) | 5 / (5 + 6) = 5/11 |
| Probability of Complement | U / (F + U) | 6 / (5 + 6) = 6/11 |
Additional Information: Odds vs. Probability
It's important to distinguish between odds and probability:
- Probability: Probability is the ratio of the number of specific outcomes to the total number of possible outcomes. It is a value between 0 and 1 (or 0% and 100%). \(P(\text{Event}) = \frac{\text{Number of specific outcomes}}{\text{Total number of outcomes}}\).
- Odds: Odds compare the number of outcomes of one type to the number of outcomes of another type (favourable vs. unfavourable). Odds are expressed as a ratio, e.g., a ∶ b. Odds can be greater than 1.
Conversion between Odds and Probability:
- If odds in favour are a ∶ b, then \(P(\text{Event}) = \frac{a}{a+b}\).
- If probability is p, then the odds in favour are \(p : (1-p)\).
- If odds against are a ∶ b, then \(P(\text{Event}) = \frac{b}{a+b}\).
- If probability is p, then the odds against are \((1-p) : p\).
Important Questions from Random Variables
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
Let $X$ and $Y$ be continuous random variables with probability density functions $P_X(x)$ and $P_Y(y)$, respectively. Further, let $Y = X^2$ and $P_X(x) = \begin{cases} 1, & x\in (0,1] \\ 0, & \text{otherwise} \end{cases}$
Which one of the following options is correct?- Two fair dice (with faces labeled 1, 2, 3, 4, 5, and 6) are rolled. Let the random variable $X$ denote the sum of the outcomes obtained.
The expectation of $X$ is __________ (rounded off to two decimal places). - Let $X = aZ + b$, where $Z$ is a standard normal random variable, and $a, b$ are two unknown constants. It is given that
$E[X] = 1$, $E[(X – E[X])Z] = –2$, $E[(X – E[X])^2] = 4$,
where $E[X]$ denotes the expectation of random variable $X$. The values of $a, b$ are: Let $Y = Z^2$, $Z = \frac{X - \mu}{\sigma}$, where $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$. The variance of $Y$ is
