Question

An experiment involved tossing two coins simultaneously. The experiment was carried out 10 times. Two heads occurred twice. What is the difference between the experimental probability of getting two heads and the theoretical probability? Write the answer in the simplest form of a fraction. (For example: 1/2 )
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1 point

Answer 1

To solve this, we first need to calculate the experimental and theoretical probabilities of getting two heads when tossing two coins.

1. **Theoretical Probability**:
When tossing two coins, the possible outcomes are:
- HH (two heads)
- HT (head and tail)
- TH (tail and head)
- TT (two tails)

There are 4 possible outcomes. The only outcome that results in two heads (HH) is one of these outcomes.

Thus, the theoretical probability of getting two heads (HH) is:
\[
P(\text{two heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{4}
\]

2. **Experimental Probability**:
In the experiment, the two coins were tossed 10 times, and two heads occurred 2 times. Therefore, the experimental probability of getting two heads is:
\[
P(\text{two heads, experimental}) = \frac{\text{Number of times two heads occurred}}{\text{Total number of tosses}} = \frac{2}{10} = \frac{1}{5}
\]

3. **Difference between Experimental and Theoretical Probability**:
Now, we need to find the difference between the experimental probability and the theoretical probability:
\[
\text{Difference} = P(\text{two heads, experimental}) - P(\text{two heads, theoretical}) = \frac{1}{5} - \frac{1}{4}
\]

To perform this subtraction, we need a common denominator. The least common multiple of 5 and 4 is 20.

Now, convert both fractions:
\[
\frac{1}{5} = \frac{4}{20}
\]
\[
\frac{1}{4} = \frac{5}{20}
\]

Now, subtract the two fractions:
\[
\frac{4}{20} - \frac{5}{20} = \frac{4 - 5}{20} = \frac{-1}{20}
\]

Thus, the difference between the experimental probability and the theoretical probability is:
\[
\boxed{-\frac{1}{20}}
\]