# Probability of getting heads and tails

Three coins are tossed, try to find the probability of getting (a) at least one head and one tail, (b) at least one head or one tail.

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There are $2^3 = 8$ possibilities. Surely, you have written down all possibilities? –  JavaMan Dec 23 '12 at 18:11

Hint: What is the chance of three heads? Three tails? How does this help? For b) what is the answer if you only flip one coin?

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a) The probability of getting a head in one toss: (1/2) The probability of getting a tail in one toss: (1/2)

$P($ at least one $head$ and at least one $tail)$

$=P($ at least one $head)+P($ at least one $tail)-P($ at least one $head$ or $tail)$

$$=1-\left(\frac12\right)^3+1-\left(\frac12\right)^3-\{1-\left(\frac02\right)^3\}$$

b)

P(at least one tail OR at least one head) = 1 - P(no tails) - P(no heads) = 1 - 1/8 - 1/8 = 6/8 = 0.75

TTT TTH THT THH HTT HHT HTH HHH

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A) H and T: Let $x_h$ and $x_t$ be the numbers of heads and tails observed and n the number of flips. We can rewrite our desired probability as: $$P(at\ least\ 1\ heads\ and\ tails) = P(x_t \ge 1\ and\ x_h \ge 1)$$ We can then use the inverse to make it easier to calculate: $$P(x_t \ge 1\ and\ x_h \ge 1) = 1 - \lnot P(x_t \ge 1\ and\ x_h \ge 1)$$ Using De Morgan's Law, this equals: $$= 1 - P(x_t = 0\ or\ x_h = 0) \\$$ Expanding the OR clause: $$= 1 - [P(x_t = 0) + P(x_h = 0) - P(x_t = 0\ and\ x_h = 0) ]$$ We can now deduce the probabilities listed above. There is only one outcome with no heads (all tails) and only one outcome with no tails (all heads). We'll divide those by the total number of outcomes. We also know that the last probability is zero, since you can't flip coins without any outcome happening. $$P(x_t = 0) = \frac{1}{2^n}\\ P(x_h = 0) = \frac{1}{2^n}$$ Therefore, the answer is (n=3): $$P(at\ least\ one\ heads\ and\ tails) = 1 - \frac{2}{2^n} = \frac{3}{4}$$
B) H or T: If this is the way you wrote it, then: $$P(x_h \ge 1\ or\ x_t \ge 1) = 1$$ Because any coin you flip must be either a heads or a tails.