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# Elementary probability exercises

We offer elementary probability exercises at the level of high school. We show the student how to use the formula of compound probabilities, how to use conditional probability, and the Bayes theorem.

## What is a random experiment?

To study phenomena, it is important to carry out experiments. In fact, depending on the situation, certain phenomena are governed by deterministic laws, and the associated experiences have only one possible result. For other phenomena, however, governed by laws, the experiments lead to several results, forming all the possible results. Thus, two similar experiments can lead to very different results. We cannot predict the exact outcome of this experiment, so it is a random experiment.

We mention that there is a deep relationship between probability and certain physical phenomena such as heat equations. Likewise where the solution is associated with a probability law called Gaussian distribution. So that it is the advanced probability that is not the subject of this article.

In this post, we only discuss elementary probability properties.

## Elementary probability exercises

Exercise: An urn contains 3 white balls and 5 red balls. We make four drawn without discount. Determine the probability of firing four red balls.

Solution: We denote by $R_i$ the event ” the i-th drawn ball is red”. Our objective is to calculate the probability of the event $A=R_1\cap R_2\cap R_3\cap R_4$.

It is extremely difficult to calculate each $\mathbb{P}(R_i)$, but the conditional probabilities $\mathbb{P}(R_i|R_1\cdots R_{i-1})$ are simple to calculate. In fact, the realization of the event $R_1\cdots R_{i-1}$ shows that the urn contains 3 white balls and $5-i+1$ red balls. Then we have a probability equal to $\frac{6-i}{9-i}$ of sorting a red ball. Hence, the formula of compound probabilities gives \begin{align*} \mathbb{P}(A)&=\mathbb{P}(R_1)\mathbb{P}(R_2|R_1)\mathbb{P}(R_3|R_1\cap R_2)\mathbb{P}(R_4|R_1\cap R_2\cap R_3)\cr &= \frac{5}{8}\times \frac{4}{7}\times \frac{3}{6}\times \frac{2}{5}\cr &= \frac{1}{14}. \end{align*}

Exercise: Steve and James train in archery. Steve reaches the target 9 times out of 10, James only 6 times out of 10. James plays two out of three. To reach the target what is the probability?

Solution: We denote by $S$ the event ” Steve plays”, $J$ the event “James plays”, and $T$ the event “the target is reached”. As $J=\overline{S}$, $(S,J)$ is a complete system of events. The formula of total probabilities implies \begin{align*} \mathbb{P}(T)&=\mathbb{P}(T|S)\mathbb{P}(S)+\mathbb{P}(T|J)\mathbb{P}(J)\cr &= \frac{9}{10}\times \frac{1}{3}+\frac{6}{10}\times \frac{2}{3}\cr &= \frac{7}{10}. \end{align*}

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