# Probability of getting heads and tails [closed]

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.

-

## closed as off-topic by Jonas Meyer, Thursday, Michael Albanese, M. Vinay, Arthur Fischer♦Sep 15 at 4:58

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Thursday, Michael Albanese, M. Vinay, Arthur Fischer
If this question can be reworded to fit the rules in the help center, please edit the question.

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?

-

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

-

Consider using the inverse to answer these questions:

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.

-