Probability of at most 1 tail? Loaded Coin Probability

Question

If someone tosses a loaded coin 40 times and the probability of each toss resulting in a head is $\frac{1}{4}$, what is the probability that the person will get at most one tail?

The answer is supposed to be $(\frac{1}{4})^{40}+(\frac{1}{4})^{39}(\frac{3}{4})\cdot 40$, however, I don't understand how this is computed.

I'm guessing the words "at most" are important to the computations, but I can't find what they are supposed to affect in terms of any formulas that may apply. Is this conditional probability?

I think maybe the two events are "exclusive or," which is why we add the probabilities together. So if the second part, $(\frac{1}{4})^{39}(\frac{3}{4})\cdot 40$, is the probability of at most one tail, why do we multiply the probability of a tail on each of the 40 tosses by the probability of a head on each of the 40 tosses and the 40 ways in which it could be arranged?

Thank you in advance to anyone who can offer some help/explanation!

Answer 1

If they get 0 tails, that means they got heads 40 consecutive times. The probability of that happening is $( \frac {1} {4})^{40}$.

If they get exactly 1 tail and 39 heads that's: $( \frac {1} {4})^{39} \cdot ( \frac 3 4 ) \cdot 40$ The $( \frac {1} {4})^{39}$ part is from the 39 heads the $( \frac 3 4 )$ is from the tail and the $40$ is because there are $40$ different places the one tail could occur.

Adding those together, the probability of exactly zero tails and the probability of exactly one tail, we get:

$( \frac {1} {4})^{40} + ( \frac {1} {4})^{39} \cdot ( \frac 3 4 ) \cdot 40$

You could also look up binomial distributions to learn more about problems like these.

Answer 2

look up Bernoulli trials: $P(k) = \binom{n}{k}p^k(1-p)^{n-k}$ where k is desired outcome and n is number of trials

you have that

• n = 40
• k = 1
• p = $\frac{1}{4}$

therefore P(1 Tail) + P(no tails) $= \binom{40}{1}(\frac{1}{4})^{39}(\frac{3}{4})^{1} + (\frac{1}{4})^{40}$