## Probability

# Conditional Probability

- The probability of occurrence of an event A, if the event B has already occurred is denoted by P(A/B) and it is defined as \tt P(A/B) = \frac{P(A \cap B)}{P(B)}, P(B) \neq 0
**Geometrical probability:**\ \tt P = \frac{Measure \ of \ the \ specified \ part \ of \ the \ region}{measure \ of \ the \ whole \ region}- If any point in the closed interval [α, β] is a sample point and any point in the closed interval [a, b] ≤ [α, β] be a favorable point for the event E. Then The probability of E is

\tt P(E) = \frac{length \ of \ [a, b]}{length \ of \ [\alpha, \beta]} - If a coin is tossed n times, then the probability of getting n heads and n-x tails = \frac{{^nc_x}}{2^{n}}
- If there are n children in a family, then the probability that there are exactly x boys (girls) is = \frac{{^nc_x}}{2^{n}}
- If an unbiased coin is tossed n times, then the probability of getting odd number of heads (tails) = \frac{1}{2}
- If a coin is tossed n times, then the probability of getting at least one head (tail) = \frac{2^{n} - 1}{2^{n}}
- If n fair dice are rolled once then the number of favarable cases to get the sum r = coefficient of x
^{2}in (x^{1}+ x^{2}+ ........ + x^{6})^{n} - Out of (2n + 1) tickets consecutively numbered three are selected at random. Then the probability that they are in A.P = \frac{3n}{4n^{2} - 1}
- If n men among whom A and B stand in a row, then the probability that there will be exactly r persons between A and B is = \frac{2(n - r - 1)}{n(n - 1)}
- If n men among whom A and B sit along a circle then the probability that there will be exactly r persons between A and B is = \begin{cases}\frac{2}{n - 1}; & {\tt if} \ r \neq \frac{n - 2}{2}\\ \frac{1}{n - 1}; & {\tt if} \ r = \frac{n - 2}{2}\end{cases}
- If n men among whom A and B sit along a circle then the probability that A and B sit together is = \frac{2}{n - 1}
- If n men among whom A and B sit along a circle, then the probability that A and B never sit together is = \frac{n - 3}{n - 1}
- If n men among whom A and B sit along a circle, then the odds against A and B sitting together is = n − 3 : 2
- When three dice are thrown, the number of ways of getting a total of r is = \begin{cases}\frac{(r - 1)(r - 2)}{2} &{\tt if}\ 3 \leq r \leq 8 \\ \frac{(19 -r)(20 - r)}{2} &{\tt if}\ 13 \leq r \leq 18 \\ 25 & {\tt if} \ r = 9 (or) 12 \\ 27 & {\tt if} \ r = 10 (or) 11\end{cases}
- When two dice are thrown, the number of ways of getting a total of r is = \begin{cases}(r - 1) &{\tt if}\ 2 \leq r \leq 7 \\ (13 - r) &{\tt if}\ 8 \leq r \leq 12 \end{cases}

### View the Topic in this video From 02:00 To 03:20

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1. P(A/B)=\frac{P(A\cap B)}{P(B)}

2. P(B/A)=\frac{P(A\cap B)}{P(A)}

3. P\left(\frac{A}{B}\right)+P\left(\frac{\overline{A}}{B}\right)=1

4. If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then

P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F)