Two identical biased coins are tossed together, and the outcome is recorded. After a large number of trials it is observed that the probability that both coins land showing heads is 0.36. What is the probability that both coins land showing tails?
Answer: 0.16
Thanks in advance to anyone who helps
though they are tossed together, the coins act independent of each other
the probability of two independent events occurring simultaneously
... is the product of their individual probabilities
p(h) * p(h) = .36 ... [p(h)]^2 = .36 ... p(h) = √.36 = .6
p(t) = 1 - p(h) = .4
probability of simultaneous tails ... [p(t)]^2 = .4^2 = .16
Well, well, well, looks like we have a coin-tossing extravaganza here! Let's dive right in, shall we?
Since the probability of getting heads on both coins is 0.36, we can say that the probability of getting tails on both coins is...wait for it...the same, right? Nope, just kidding!
But fear not, my friend, because we can still figure this out! If the coins are identical and unbiased, that means the probability of getting heads on one coin is 0.5 (50%) and the probability of getting tails is also 0.5.
Now, let's use a bit of probability magic. If the probability of getting heads on both coins is 0.36, then the probability of getting tails on both coins should be the same. So, we can subtract 0.36 from 1 (since there are only two possible outcomes: heads or tails) to get the answer.
Drumroll, please! The probability that both coins land showing tails is 0.64. Ta-da! And just like that, we've solved the mystery of the double-tailed coins!
I hope this answers your question, and remember, laughter is the best currency!
To solve this problem, let's first assign some variables:
Let A represent the event of both coins showing heads.
Let B represent the event of both coins showing tails.
We are given that the probability of event A (P(A)) is 0.36. Since there are only two possible outcomes (heads or tails) for each coin and the two coins are identical, we can assume that the probability of event B (P(B)) is the same as the probability of event A.
The sum of the probabilities of all possible outcomes must be equal to 1. Since there are only two possible outcomes (heads or tails), we have:
P(A) + P(B) = 1
Substituting the given probability of event A:
0.36 + P(B) = 1
Rearranging the equation to solve for P(B):
P(B) = 1 - 0.36
P(B) = 0.64
Therefore, the probability that both coins land showing tails (P(B)) is 0.64 or 0.16 when rounded to two decimal places.
To find the probability that both coins land showing tails, we can start by finding the probability that both coins land showing heads. Let's call this probability "p":
p = 0.36
Since there are only two possible outcomes for each coin toss (heads or tails), the probability of getting tails on a single coin toss can be represented as (1 - p), where (1 - p) is the complement of p.
Now, the probability of both coins landing with tails can be calculated by multiplying the probability of getting tails on a single coin toss twice, since the coins are tossed together. Therefore, the probability of both coins landing showing tails can be calculated as follows:
Probability of getting tails on a single coin toss = (1 - p)
Probability of both coins landing tails = (1 - p) * (1 - p)
Substituting the given p = 0.36, we can calculate the probability:
Probability of both coins landing tails = (1 - 0.36) * (1 - 0.36)
= 0.64 * 0.64
= 0.4096
Therefore, the probability that both coins land showing tails is 0.4096, which is approximately 0.41.
It's important to note that this calculation assumes that the two coin tosses are independent and the biases of the two coins are the same.