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.



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


$P = \text{HTT or THT or TTH}$

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

$P = \dfrac{3}{8}$   ←   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 1st toss 2nd toss 3rd 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.


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