Suppose you decided to keep flipping a coin until it came up tails. Which of these is most likely?
a.flipping the coin 11 times
b.flipping the coin 12 times
c.flipping the coin 14 times
d.flipping the coin 13 times
P(n Heads,Tail) = (1/2)n * 1/2 = (1/2)^(n+1)
Which answer is it though
To determine which option is most likely, we need to understand the probability of flipping a coin and the concept of expected value.
When flipping a fair coin, there are two possible outcomes: heads or tails, each with a 50% chance. Suppose you flip a coin and record the result. If it comes up heads, you flip it again, and so on until you flip tails.
Let's break down the options:
a. Flipping the coin 11 times: The probability of flipping tails on the first flip is 0.5. If it doesn't come up tails, you continue flipping, and the probability of flipping tails on the second flip is also 0.5. This process continues until you reach the 11th flip. The probability of flipping tails on the 11th flip is (0.5)^11 ≈ 0.000488 or 0.0488%.
b. Flipping the coin 12 times: Using the same reasoning as above, the probability of flipping tails on the 12th flip is (0.5)^12 ≈ 0.000244 or 0.0244%.
c. Flipping the coin 14 times: Again, the probability of flipping tails on the 14th flip is (0.5)^14 ≈ 0.000122 or 0.0122%.
d. Flipping the coin 13 times: Similar to the previous examples, the probability of flipping tails on the 13th flip is (0.5)^13 ≈ 0.000244 or 0.0244%.
Comparing the probabilities, option a has the highest likelihood of occurring. So the most likely option is flipping the coin 11 times before it comes up tails.
all are equal.
if one the tenth, to flip again, pr(tail)=1/2
if you didn't get it on the tenth, the pr(tail)=1/2 and so on.
no matter how long previous flips have gone, the prob of a tail on one more flip is the same, 1/2.