Mathematics 10 — Exercise 12.2 · Q9

Full question

Two fair coins are tossed. Find the probability of obtaining at least one head or two heads.

Step-by-step solution

1. All outcomes are equally likely. Count the complete sample space.

   $$\mathrm{n(S)}=4$$

2. Select exactly the outcomes satisfying the written condition. These are listed below.

   $$\mathrm{n(E)}=3$$

3. Probability=favourable outcomes / total equally likely outcomes.

   $$\frac{3}{4}$$

4. Calculate this operation before combining the remaining terms.

   $$3\times 1=3$$

5. Calculate this operation before combining the remaining terms.

   $$1\times 4=4$$

6. For multiplication of fractions, multiply the numerators together and the denominators together.

   $$\frac{3\times 1}{1\times 4}$$

7. Calculate these two products separately.

   $$\frac{3}{4}$$

8. For multiplication of fractions, multiply the numerators together and the denominators together.

   $$\frac{1\times 1}{1\times 4}$$

9. Calculate these two products separately.

   $$\frac{1}{4}$$

10. Calculate this operation before combining the remaining terms.

    $$1\times \frac{1}{4}=\frac{1}{4}$$

11. Calculate this operation before combining the remaining terms.

    $$3\times \frac{1}{4}=\frac{3}{4}$$

12. Reduce the fraction by dividing numerator and denominator by their greatest common divisor.

    $$\frac{3}{4}$$

13. Two heads is already included in at least one head; do not count HH twice.

Answer

The final answer is shown below.