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In this problem, it is said that a box contains 10 coins, 5 coins are 2 headed, 3 coins are 2 -tailed, and 2 are fair coins.
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A coin is chosen at random and tossed.
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In the first sub -part, we need to find the probability that a head appears.
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Now, this will be the sum of the probabilities of the different cases.
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Now, the first case is that a coin is 2 -headed.
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Now, the probability of choosing a 2 -headed coin at random will be 5 by 10, since 5 out of the 10 coins are 2 -headed.
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With this, we multiply the probability of getting a heads.
00:32
Now, if it is a two -headed coin, that means no matter what, we will always get a heads.
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So that means that the probability of a head appearing will be one.
00:40
So we add with this the probability that we get a heads if we have a two -tailed coin.
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Now, the probability of choosing a two -tailed coin is three by ten, since there are three out of ten coins which are two -tailed.
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And we multiply with this the probability of getting a heads.
00:53
Since it is a two -tailed coin, that means we will never get a heads, and the probability of getting a heads will be zero.
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With this, we add the probability of getting a heads if we have a fair coin.
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Now, the probability of randomly choosing a fair coin will be 2 by 10 since there are two fair points.
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And we multiply this with the probability of getting a heads.
01:11
Since it is a fair coin, there is one favorable outcome of getting a heads...