Question

3.5 coin flips. if you flip a fair coin 10 times, what is the probability of (a) getting all tails? (b) getting all heads? (c) getting at least one tails? (a) 0.5^10 = 0.00098. (b) 0.5^10 = 0.00098. (c) p(at least one tails) = 1 - p(no tails) = 1 - (0.5^10) ≈ 1 - 0.001 = 0.999.

Explanation:

Step1: Probability of single - coin flip

The probability of getting a head or a tail in a single fair - coin flip is $p = 0.5$.

Step2: Probability of all tails

The probability of getting all tails in $n = 10$ independent coin flips is given by the multiplication rule for independent events. Since the events are independent, the probability of getting all tails is $P(\text{all tails})=(0.5)^{10}=0.00098$.

Step3: Probability of all heads

The probability of getting all heads in $n = 10$ independent coin flips is also given by the multiplication rule for independent events. So $P(\text{all heads})=(0.5)^{10}=0.00098$.

Step4: Probability of at least one tail

The probability of getting at least one tail is the complement of the probability of getting no tails (i.e., all heads). Let $P(X\geq1\text{ tail}) = 1 - P(X = 0\text{ tails})$. Since $P(X = 0\text{ tails})=(0.5)^{10}$, then $P(X\geq1\text{ tail})=1 - 0.00098 = 0.99902\approx0.999$.

Answer:

(a) $0.00098$
(b) $0.00098$
(c) $0.999$