A fair coin is tossed 4 times. Given that each of the first 3 tosses land tails up, what is the probability that all 4 tosses land tails up?
i think ...
isn't it
3/4?
1/2.. Same as a single coin toss. The odds of each toss are independent of what went before.
oh yeah duh haha
it doesn't matter what was before there're two choices haha i remember now :]
hey answer my question!!
I did answer your question
To find the probability that all 4 tosses land tails up, given that each of the first 3 tosses land tails up, we can use the concept of conditional probability.
The probability of an event A occurring, given that event B has occurred, is denoted as P(A|B), and it is calculated using the formula:
P(A|B) = P(A ∩ B) / P(B)
Here, A represents the event "all 4 tosses land tails up" and B represents the event "each of the first 3 tosses land tails up".
Now, let's find the probabilities required for the formula:
P(A ∩ B) denotes the probability that both events A and B occur simultaneously. In this case, it is the probability that all 4 tosses land tails up, which is (1/2)^4 = 1/16, since there are 2 equally likely outcomes (tails or heads) for each toss.
P(B) denotes the probability of event B occurring. Given that each of the first 3 tosses land tails up, the probability of getting tails on each individual toss is (1/2)^3 = 1/8.
Substituting these values into the formula, we have:
P(A|B) = (1/16) / (1/8) = 1/2
Therefore, the probability that all 4 tosses land tails up, given that each of the first 3 tosses land tails up, is 1/2 or 50%.