A probability model for some experiment is a distribution of probability that will predict the relative frequency of every outcome that is possible when the experiment is performed for a large number of times.

This also refers to usually as theoretical probability.

Related Calculators | |

Calculation of Probability | Binomial Distribution Probability Calculator |

Binomial Probability Calculator | Coin Toss Probability Calculator |

Some Important Points to Remember:

When we flip a fair coin and then observe which side faces up of the coin, then both heads and tails are to come up equally likely. This can be seen as the probability distribution easily as these numbers add up to 1 and also lie between 0 and 1. The sample space $S$ = {$H, T$} and the probability of occurrence of heads that is $P (H)$ = 0.5 and also of tails $P (T)$ = 0.5.

In an unfair coin the sample space, $S$ = {$H, T$}. One can choose any non zero and non negative values for both $P (H)$ and $P (T)$ with the sum of $P (H)$ and $P (T)$ to be 1 as the coin is unfair.

**In case of a dice which is fair it is expected to roll every number on the die one sixth of time. So it is also a probability distribution example. **

The sample space, $S$ = {1, 2, 3, 4, 5, 6} and the probability of occurrence of numbers, $P$ (1) = $P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = 1/6.

In case of rolling of two dice, there are in total of 36 outcomes that too if the dice are fairly distinguishable. So the sample space here is given as follows:

$S$ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) and (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) and (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) and (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) and (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

The probability of each outcome will be $\frac{1}{36}$.

**SOLVED EXAMPLES**

**Example 1:** In we are rolling two fair dices and let $E$ be the event that 6 is the sum of the numbers that face up. Find the probability of the event E.

$E$ in this case, $E$ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}.

Now by using the properties of probability distribution, the probability of the event $E$ of having 6 as the sum of the numbers that will face up:

$P (E)$ = $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ = $\frac{5}{36}$.

**Example 2:** If we toss a coin three times and E is the occurrence of at least two heads. Find the probability of the event E.

Here $S$ = {$HHH, HHT, HTH, HTT, THT, THH, TTH, TTT$}.

That is number of outcomes $n (S)$ = 8.

Now, occurrence of at least two heads implies, $E$ = {$HHT, HTH, THH, HHH$}.

This will imply that $n (E)$ = 4.

**So the probability of occurrence of at least two heads modeled with probability distribution can be given by:**

$P (E)$ = $\frac{n (E)}{n (S)}$ = $\frac{4}{8}$ = $\frac{1}{2}$.

**Example 3: **Let us assume that there are 5 colored marbles in a bowl. One of them is red, one is blue, one is yellow, one is green and one is purple. Of one of the marble is to be picked randomly from the bowl, the possible outcomes can be, S = {red, blue, yellow, green, purple}. So the probability distribution modeled value assign to each will be equal to $\frac{1}{5}$.

When we flip a fair coin and then observe which side faces up of the coin, then both heads and tails are to come up equally likely. This can be seen as the probability distribution easily as these numbers add up to 1 and also lie between 0 and 1. The sample space $S$ = {$H, T$} and the probability of occurrence of heads that is $P (H)$ = 0.5 and also of tails $P (T)$ = 0.5.

In an unfair coin the sample space, $S$ = {$H, T$}. One can choose any non zero and non negative values for both $P (H)$ and $P (T)$ with the sum of $P (H)$ and $P (T)$ to be 1 as the coin is unfair.

The sample space, $S$ = {1, 2, 3, 4, 5, 6} and the probability of occurrence of numbers, $P$ (1) = $P$ (2) = $P$ (3) = $P$ (4) = $P$ (5) = $P$ (6) = 1/6.

In case of rolling of two dice, there are in total of 36 outcomes that too if the dice are fairly distinguishable. So the sample space here is given as follows:

$S$ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) and (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) and (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) and (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) and (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

The probability of each outcome will be $\frac{1}{36}$.

$E$ in this case, $E$ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}.

Now by using the properties of probability distribution, the probability of the event $E$ of having 6 as the sum of the numbers that will face up:

$P (E)$ = $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ + $\frac{1}{36}$ = $\frac{5}{36}$.

Here $S$ = {$HHH, HHT, HTH, HTT, THT, THH, TTH, TTT$}.

That is number of outcomes $n (S)$ = 8.

Now, occurrence of at least two heads implies, $E$ = {$HHT, HTH, THH, HHH$}.

This will imply that $n (E)$ = 4.

$P (E)$ = $\frac{n (E)}{n (S)}$ = $\frac{4}{8}$ = $\frac{1}{2}$.

Related Topics | |

Math Help Online | Online Math Tutor |