die and coin
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Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
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Probabilities... Hmm... Strange question, but interesting nonetheless.
From 1 to 6 you get two multiples of 3: 3 and 6. You will need to get 3 and 6 for your first toss. Then it gets a bit more complicated (or at least that's what I understand from if any other number comes). If you got a 6 the first time, now you need a 3 and vice versa. The coin toss will be simpler, since there's a fixed 50% chance of getting tails.
So two cases, each with the same probability. Getting a 3 the first time is 1/6, while getting a 6 the second time is again 1/6. Getting the tails will be 1/2. The probability for this event will be 1/6 * 1/6 * 1/2 = 1/72. Use the same reasoning to conclude that getting a 6 the first time and a 3 the second time yields the same probability.
Add them up to get 1/72 + 1/72 = 2/72 = 1/36.
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Answer is 0.DanaJ wrote:
Probabilities... Hmm... Strange question, but interesting nonetheless.
From 1 to 6 you get two multiples of 3: 3 and 6. You will need to get 3 and 6 for your first toss. Then it gets a bit more complicated (or at least that's what I understand from if any other number comes). If you got a 6 the first time, now you need a 3 and vice versa. The coin toss will be simpler, since there's a fixed 50% chance of getting tails.
So two cases, each with the same probability. Getting a 3 the first time is 1/6, while getting a 6 the second time is again 1/6. Getting the tails will be 1/2. The probability for this event will be 1/6 * 1/6 * 1/2 = 1/72. Use the same reasoning to conclude that getting a 6 the first time and a 3 the second time yields the same probability.
Add them up to get 1/72 + 1/72 = 2/72 = 1/36.
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