Example Question - conditional probability
Here are examples of questions we've helped users solve.
Probability of Picking Even Numbers Successively
To solve this problem, we need to calculate the probability of two events happening one after another: 1. Picking an even number on the first card. 2. Picking an even number on the second card, without replacing the first card. Let's break it down: 1. For the first event, we see there are 4 even numbers in the set of cards (2, 4, 6, and the hidden card which logically would be 8 to complete the sequence). Therefore, there are 4 even cards out of the total 7 cards. The probability of picking an even number on the first try is: P(even first) = Number of even cards / Total number of cards = 4/7 2. For the second event, since we picked an even card first and did not replace it, there are now 3 even cards left out of 6 total cards. The probability of picking an even number on the second try, given that the first card drawn was even, is: P(even second | even first) = Remaining even cards / Remaining total cards = 3/6 Now, for the overall probability of both events happening one after the other, we multiply the two probabilities: P(even first and even second) = P(even first) * P(even second | even first) Plugging in the values, we get: P(even first and even second) = (4/7) * (3/6) Simplifying the fraction, we get: P(even first and even second) = (4/7) * (1/2) = 4/14 = 2/7 So the probability of picking an even number and then picking another even number without replacement is 2/7.
Probability of Selecting a Double-Headed Coin
The image contains a handwritten probability question that reads as follows: "I have in my pocket 10 coins - 9 of them are ordinary coins with equal chances of coming head and tail when tossed and the 10th has heads on both sides. a) If I take one of the coins at random from my pocket (without looking) and toss it, what is the probability that it is the coin with 2 heads? b) If I tossed a coin and it comes up head, what is the probability that it is the coin with 2 heads? c) If I toss a coin a further two times and it comes up head both times, what is the probability that it is a 2 head coin rather than an ordinary one?" Let's solve each part of the question: a) Since you have 10 coins with only one having 2 heads, the probability of picking the double-headed coin at random is simply the ratio of double-headed coins to the total number of coins. Hence, the probability is 1/10. b) Here we are given additional information that the coin we selected came up heads. This changes the scenario because we must update our probabilities based on this new information. This is a conditional probability problem, which we can approach using Bayes' theorem. Let H denote the event that the coin toss results in heads and D denote the event that the chosen coin is the double-headed coin. We want to find P(D|H), the probability that the coin is double-headed given it landed on heads. P(D|H) = P(H|D) * P(D) / P(H) We know that: P(H|D) = 1 (because the double-headed coin always lands on heads) P(D) = 1/10 (from part a) P(H) = P(H and D) + P(H and not D) = P(H|D)*P(D) + P(H|not D)*P(not D) = 1 * 1/10 + 1/2 * 9/10 (since an ordinary coin has a 1/2 chance of landing on heads) = 1/10 + 9/20 = 2/20 + 9/20 = 11/20 Now we can find P(D|H): P(D|H) = (1 * 1/10) / (11/20) P(D|H) = 1/10 * 20/11 P(D|H) = 2/11 So the probability that the coin is the double-headed one given that it came up heads is 2/11. c) We need to update our probabilities again with the new information: the coin has come up as heads two more times. The probability of choosing the double-headed coin and getting heads three times in a row (the one initial time plus two more times) can be updated in the same manner as in part b. P(D|HHH) = P(HHH|D) * P(D) / P(HHH) We already know P(D) and P(HHH|D), but P(HHH) is new: P(HHH) = P(HHH and D) + P(HHH and not D) = P(HHH|D)*P(D) + P(HHH|not D)*P(not D) = 1 * 1/10 + (1/2)^3 * 9/10 (since an ordinary coin has a 1/2 chance of landing on heads each time) = 1/10 + 1/8 * 9/10 = 1/10 + 9/80 = 8/80 + 9/80 = 17/80 Now we can find P(D|HHH): P(D|HHH) = (1 * 1/10) / (17/80) P(D|HHH) = 1/10 * 80/17 P(D|HHH) = 8/17 The probability that it is a double-headed coin rather than an ordinary one when it has landed heads three times is 8/17.
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