Example Question - successive events
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.
Calculating Probability of Events Happening Successively
To solve this probability question, we need to calculate the probability of two independent events happening one after the other: 1. Picking an even number. 2. Picking an 8. Since the card is put back after the first pick, the two events are independent, and we can calculate the probability of each event separately and then multiply them. First, let's find the probability of picking an even number from the cards shown. There are three cards, and only one of them is even (the 8). So the probability of picking an even number is: P(even) = Number of even cards / Total number of cards = 1/3 Next, let's find the probability of picking an 8. Since the card is put back after the first draw, the probability remains the same for picking an 8: P(8) = Number of cards with 8 / Total number of cards = 1/3 Now we multiply the probabilities of the two independent events: P(even, then 8) = P(even) * P(8) = (1/3) * (1/3) = 1/9 So, the probability of picking an even number and then picking an 8, with replacement, is 1/9.
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