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Nina [5.8K]
3 years ago
12

A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If

the coin turns up heads, what is the probability that it is the two-headed coin? Is the answer 1/3? Was the answer intuitive?
Mathematics
1 answer:
Blababa [14]3 years ago
5 0

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

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A geometric series has second term 375 and fifth term 81 . find the sum to infinity of series .
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<u>Step-by-step explanation:</u>

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First, let's find the ratio (r). There are three multiple from 375 to 81.

375r^3=81\\\\r^3=\dfrac{81}{375}\\\\\\r^3=\dfrac{27}{125}\qquad \leftarrow simplied\\\\\\\sqrt[3]{r^3} =\sqrt[3]{\dfrac{27}{125}}\\ \\\\r=\dfrac{3}{5}

Next, let's find a₁

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Lastly, Use the Infinite Geometric Sum Formula to find the sum:

S_{\infty}=\dfrac{a_1}{1-r}\\\\\\.\quad =\dfrac{625}{1-\frac{3}{5}}\\\\\\.\quad =\dfrac{625}{\frac{2}{5}}\\\\\\.\quad = \dfrac{625(5)}{2}\\\\\\.\quad = \large\boxed{\dfrac{3125}{2}}

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