In this post, we offer exercises on probability for high school. Our objective is to show the student how to use the formula of compound probabilities, how to use conditional probability and the Bayes theorem.

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

### Exercises on probability for high school

**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, if the event $R_1\cdots R_{i-1}$ is realised, the urn contain 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. What is the probability that the target is reached?

**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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