# Probability Problems Involving Coins

When you toss a coin, the outcome can either be *head* or *tail*. If the coin is so balanced that these two outcomes are equally likely to occur, then the probability that the outcome is head is 1/2, and the probability that the outcome is tail is also 1/2.

**Example**

A coin is tossed three times. What is the probability that the head will appear only once?

**Solution**

$P = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} ~ + ~ \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} ~ + ~ \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}$

$P = \dfrac{3}{8}$ ← *answer*

Or you can solve this using the *binomial distribution* formula

$P(x) = {^n}C_x \, p^x \, q^{n - x}$

$P(1 \text{ is H}) = {^3}C_1 \, H^1 \, T^{3 - 1}$

$P = {^3}C_1 \cdot 0.5^1 \cdot 0.5^2$

$P = 0.375$ ← *answer*

The best way however to visualize the situation is by tabulation.

Outcome | 1^{st} toss |
2^{nd} toss |
3^{rd} toss |
---|---|---|---|

1 | H | H | H |

2 | H | H | T |

3 | H | T | H |

4 | T | H | H |

5 | H | T | T |

6 | T | H | T |

7 | T | T | H |

8 | T | T | T |

A total of 8 possible outcomes and 3 outcomes containing exactly one H. Rows 5, 6, and 7 shows the condition of one H.

Hence, the probability for H to appear only once is

$P = \dfrac{3}{8}$ ← *answer*

Observe that the same outcomes get tabulated if you have to toss 3 coins once.