Question - Probability of Picking Even Numbers Successively
mathotherconditional probabilityeven numbers probabilitysuccessive eventspicking cards without replacement
Solution:
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 = 4/72. 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 = 3/6Now, 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 = 2/7So the probability of picking an even number and then picking another even number without replacement is 2/7.
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