What are the odds of having a coin land tails up on 4our consecutive tosses of the coin. I took (1/2)^4 and got 1/16 but my teacher counted it wrong is 1/16th right?
"odds" do not have the same meaning as probability
in general the odds in favour of an event is the ratio of the probability of that event happening to the probability of the event not happening.
so you were partly right, the probability of 4 cosecutive tails is 1/15
so the prob. of not getting 4 consecutive tails is 15/16
and the odds in favour of 4 consecutive tails is 1/16:15/16
= 1:15
"so you were partly right, the probability of 4 cosecutive tails is 1/15"
should have said:
so you were partly right, the probability of 4 cosecutive tails is 1/16
Yes, your calculation is correct. The probability of landing tails up on a single coin toss is 1/2, assuming the coin is fair. Since the coin tosses are independent events, you can multiply the probabilities together to calculate the probability of all four coin tosses landing tails up.
Using the formula (1/2)^4:
(1/2) * (1/2) * (1/2) * (1/2) = 1/16
Therefore, the probability of having a coin land tails up on four consecutive tosses is indeed 1/16. It appears that there may have been an error in counting by your teacher.
You were correct in calculating the probability of getting tails on a single toss of a fair coin, which is \( \frac{1}{2} \). However, to find the probability of getting tails on four consecutive tosses, you need to multiply the individual probabilities together.
So, the probability of getting tails on the first toss is \( \frac{1}{2} \), the probability of getting tails on the second toss is also \( \frac{1}{2} \), and so on. Therefore, the probability of getting tails on four consecutive tosses is \( \left(\frac{1}{2}\right)^4 \), which is indeed \( \frac{1}{16} \).
It's worth noting that probabilities can be expressed as fractions, decimals, or percentages. In this case, \( \frac{1}{16} \) is the fraction form.