The two-way table shows information about the pupils in a primary school.

\begin{tabular}{|c|c|c|c|}
\hline & Boys & Girls & Total \\
\hline Can count & 42 & 58 & 100 \\
\hline Cannot count & 11 & 9 & 20 \\
\hline Total & 53 & 67 & 120 \\
\hline
\end{tabular}

a) How many boys can count?

b) What is the probability that a randomly selected pupil is a boy and can count?



Answer :

Sure, let's break down the solution to each part of the question using the given information in the two-way table.

### Part (a): How many boys can count?

From the two-way table, we can directly see the number of boys who can count:
[tex]\[ \begin{tabular}{|c|c|c|c|} \hline & Boys & Girls & Total \\ \hline Can count & 42 & 58 & 100 \\ \hline Cannot count & 11 & 9 & 20 \\ \hline Total & 53 & 67 & 120 \\ \hline \end{tabular} \][/tex]

In the row labeled "Can count" under the column labeled "Boys," the number is 42. Therefore, the answer to part (a) is:
[tex]\[ \text{Number of boys who can count} = 42 \][/tex]

### Part (b): What is the probability that any pupil is a boy and can count?

To find the probability that a randomly selected pupil is a boy and can count, we need to consider the number of boys who can count and the total number of pupils.

We already know from part (a) that the number of boys who can count is 42. Additionally, the total number of pupils is given as 120 (sum of all entries in the "Total" row of the two-way table).

The probability [tex]\( P \)[/tex] can be computed as the ratio of the number of favorable outcomes (boys who can count) to the total number of outcomes (total number of pupils):

[tex]\[ P(\text{Boy and can count}) = \frac{\text{Number of boys who can count}}{\text{Total number of pupils}} = \frac{42}{120} \][/tex]

Now, converting this fraction to a decimal:

[tex]\[ P(\text{Boy and can count}) = \frac{42}{120} = 0.35 \][/tex]

Therefore, the probability that a randomly selected pupil is a boy and can count is:
[tex]\[ \text{Probability} = 0.35 \][/tex]

### Final Answers:

a) Number of boys who can count: [tex]\( 42 \)[/tex]

b) Probability that any pupil is a boy and can count: [tex]\( 0.35 \)[/tex]