A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times
a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?
b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.
Can someone please help me solve it using sequences of events?
Thanks
a) Let A be the event that there is at least one head in the three tosses, and B be the event that there are at least two heads.
We want to compute P(B|A), the probability of B given A. By the definition of conditional probability, we have P(B|A) = P(A∩B) / P(A).
First, let's compute the probabilities of the complementary events A' and B', which are "no heads in the three tosses" and "fewer than two heads", respectively.
For A': the only way to get no heads is to have the coin land tails three times. This happens with probability (1/3)^3 = 1/27. Since P(A) + P(A') = 1, we get P(A) = 26/27.
For B': there can be either no heads or exactly one head. We already computed the probability of no heads, which is 1/27. To get exactly one head, there are three scenarios: H-T-T, T-H-T, or T-T-H. The probability of each of these scenarios is (2/3)*(1/3)*(1/3) = 2/27. In total, P(B') = P(no heads) + P(exactly one head) = 1/27 + 3*(2/27) = 1/3. Since P(B) + P(B') = 1, we get P(B) = 2/3.
Now we can compute the intersection of A and B, A∩B. We know that B implies at least two heads, so A∩B is just the event "at least two heads". We know that P(at least two heads) = P(B) = 2/3.
Finally, we can compute the conditional probability P(B|A) = P(A∩B) / P(A) = (2/3) / (26/27) = (2/3)*(27/26) = 9/13.
b) Let C be the event of exactly one head. To find P(C|A), notice that C is the complement of B given A, since if there is at least one head and there are not at least two heads, then there must be exactly one head. Thus, P(C|A) = 1 - P(B|A) = 1 - 9/13 = 4/13.
To solve this problem using sequences of events, we need to break it down step by step. Let's analyze each question separately:
a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?
To find the probability that there were at least two heads, we need to consider all possible outcomes where there are at least two heads out of the three tosses, given that there was at least one head.
Step 1: Determine the sample space for the three tosses. Since each toss has two possible outcomes (heads or tails), the sample space contains 2^3 = 8 equally likely outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
Step 2: Identify the outcomes that have at least one head. In this case, all the outcomes except TTT have at least one head: {HHH, HHT, HTH, HTT, THH, THT, TTH}.
Step 3: Determine the outcomes that have at least two heads. From the outcomes with at least one head, we can see that the following outcomes have at least two heads: {HHH, HHT, HTH, THH}.
Step 4: Calculate the probability that there were at least two heads, given that there was at least one head. The probability is calculated by dividing the number of favorable outcomes (outcomes with at least two heads) by the number of possible outcomes (outcomes with at least one head). So in this case, the probability is 4 (outcomes with at least two heads) divided by 7 (outcomes with at least one head). Therefore, the probability is 4/7.
b) Use your answer in a) to find the probability that there was exactly one head, given that there was at least one head in the three tosses.
To find the probability that there was exactly one head, given that there was at least one head, we can subtract the probability of having at least two heads (found in part a) from the probability of having at least one head.
Step 1: Calculate the probability of having at least one head in the three tosses. From the outcomes with at least one head (found in part a), we know that there are 7 outcomes with at least one head out of the 8 possible outcomes. Therefore, the probability of having at least one head is 7/8.
Step 2: Subtract the probability of having at least two heads (found in part a) from the probability of having at least one head. So the probability of having exactly one head is 7/8 (probability of having at least one head) - 4/7 (probability of having at least two heads). Perform the calculation to find the probability of exactly one head.
I hope this helps you solve the problem using sequences of events! Let me know if you have any further questions.
Sure! Let's solve the problem step-by-step using sequences of events.
a) To find the probability that there were at least two heads, given that there was at least one head in the three tosses, we can use conditional probability. Let's denote the events as follows:
A = at least one head in three tosses
B = at least two heads in three tosses
We need to find P(B|A), the probability of B given A. Using the definition of conditional probability, we have:
P(B|A) = P(A and B) / P(A)
To calculate P(A and B), we can use the probability of the intersection of the two events. Since B implies A (if there are at least two heads, there is at least one head), we have:
P(A and B) = P(B)
To calculate P(B), we need to consider all the possible outcomes that satisfy B. These outcomes are HHT, HTH, and THH (where H represents a head and T represents a tail). Since the probability of a head is 2/3 and the probability of a tail is 1/3, we have:
P(B) = P(HHT) + P(HTH) + P(THH)
= (2/3 * 2/3 * 1/3) + (2/3 * 1/3 * 2/3) + (1/3 * 2/3 * 2/3)
= 4/27 + 4/27 + 4/27
= 12/27.
Now, let's calculate P(A), the probability of at least one head in three tosses. To find this probability, we can use the complement rule. The complement of at least one head is no heads, i.e., all three tosses resulting in tails. Thus,
P(A) = 1 - P(no heads)
= 1 - (1/3 * 1/3 * 1/3)
= 1 - 1/27
= 26/27.
Finally, we can calculate P(B|A) using the formula:
P(B|A) = P(B) / P(A)
= (12/27) / (26/27)
= 12/26
= 6/13.
Therefore, the probability that there were at least two heads, given that there was at least one head in the three tosses, is 6/13.
b) To find the probability that there was exactly one head, given that there was at least one head in the three tosses, we can use the complement rule. Let's define the event C as exactly one head in three tosses.
P(C|A) = 1 - P(not C|A)
Now, let's find P(not C|A), the probability of not having exactly one head, given that there was at least one head in the three tosses. This is equal to the probability of having no heads or having at least two heads. We already know that the probability of at least two heads is 6/13.
P(not C|A) = P(no heads or at least two heads) = P(A) - P(C|A)
= 26/27 - P(C|A).
Since the probability of having at least two heads is equal to 6/13, we have:
P(not C|A) = 26/27 - 6/13
= (26 * 13 - 6 * 27) / (27 * 13)
= 304/351.
Finally, we can calculate P(C|A) using the complement rule:
P(C|A) = 1 - P(not C|A)
= 1 - 304/351
= (351 - 304) / 351
= 47/351.
Therefore, the probability that there was exactly one head, given that there was at least one head in the three tosses, is 47/351.