Question

In the class of 30 children,16 are girls,4 wear glasses and 3 are left-handed. A child is chosen at random from the class. What is the probability that this child is:
(a) A girl
(b) Right-handed
(c) Wearing glasses

Answer 1

To solve for the probabilities, we need to use the basic probability formula:

\[
P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}
\]

In our case, the total number of outcomes is 30 (the total number of children in the class).

### (a) Probability that the child is a girl:

Number of girls = 16

\[
P(\text{Girl}) = \frac{\text{Number of girls}}{\text{Total number of children}} = \frac{16}{30} = \frac{8}{15} \approx 0.5333
\]

### (b) Probability that the child is right-handed:

To find the number of right-handed children, we need to calculate it as follows:

Total children = 30

Number of left-handed children = 3

This means the number of right-handed children = Total children - Left-handed children = 30 - 3 = 27.

\[
P(\text{Right-handed}) = \frac{\text{Number of right-handed children}}{\text{Total number of children}} = \frac{27}{30} = \frac{9}{10} = 0.9
\]

### (c) Probability that the child is wearing glasses:

Number of children who wear glasses = 4

\[
P(\text{Wearing glasses}) = \frac{\text{Number of children wearing glasses}}{\text{Total number of children}} = \frac{4}{30} = \frac{2}{15} \approx 0.1333
\]

### Summary:
- Probability that the child is a girl: \(\frac{8}{15} \approx 0.5333\)
- Probability that the child is right-handed: \(\frac{9}{10} = 0.9\)
- Probability that the child is wearing glasses: \(\frac{2}{15} \approx 0.1333\)