If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
6 ∶ 5
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.
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:
We are given that the odds in favour of event A are 5 ∶ 6.
This can be interpreted as:
The total number of possible outcomes is the sum of favourable and unfavourable outcomes:
\(\text{Total Outcomes} = \text{Favourable Outcomes} + \text{Unfavourable Outcomes}\)
\(\text{Total Outcomes} = 5 + 6 = 11\)
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:
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 |
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 |
It's important to distinguish between odds and probability:
Conversion between Odds and Probability:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
| List - I | List - II |
| (a) \(f_{X}(x)\) | (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) |
| (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) | (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) |
| (c) \(F_{X}(a)\) | (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) |
| (d) \(f_{Y}(y)\) | (iv) 1 |
Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :