Question

you toss a fair coin 4 times. what is the probability that (round to 4 decimal places) a) you get all heads? enter an integer or decimal number, accurate to at least 4 decimal places more.. b) you get at least one tail?

Explanation:

Step1: Calculate total possible outcomes

Each coin - toss has 2 possible outcomes. For 4 coin - tosses, by the multiplication principle, the total number of possible outcomes is $2\times2\times2\times2 = 2^4=16$.

Step2: Calculate probability of all heads

The event of getting all heads is 1 specific outcome. The probability of an event $E$ is $P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So the probability of getting all heads is $\frac{1}{16}=0.0625$.

Step3: Calculate probability of at least one tail

The complement of the event “getting at least one tail” is the event “getting all heads”. Let $A$ be the event of getting at least one tail and $B$ be the event of getting all heads. Then $P(A)=1 - P(B)$. Since $P(B) = 0.0625$, then $P(A)=1 - 0.0625 = 0.9375$.

Answer:

a) 0.0625
b) 0.9375