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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
Will the value of A be (20/30 * 10/30) or (20/30 * 10/30* 2) i.e P(fair ball first and unfair ball after) + P(unfair first and then fair after)

Does order matter here?
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
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Agreed with Pranab01

Probability of exactly 1 fair and 1 not fair coin can be done in either way he mentioned above

Only thing I would add is to make sure the denominator is 30 in this instance for both (Fair and non fair) specifically because of the wording "with replacement".
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
pranab223 wrote:
Carcass wrote:
A bag has 30 different coins, of which 20 are fair. 2 coins are selected with replacement from the bag.


Quantity A
Quantity B
Probability of selecting exactly one fair coin
\(\frac{20}{30} \times \frac{10}{29}\)


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

OA is not provided



Here,

there are 2 possibilities of selecting exactly one fair coin

Possibility 1:

One fair coin in first selection and Not a fair coin in second selection

Then the probability = \(\frac{20}{30} * \frac{10}{30} = \frac{2}{9}\)

Possibility 2:

Not a fair coin in first selection and a Fair coin in second selection

Then the probability = \(\frac{10}{30} * \frac{20}{30} = \frac{2}{9}\)

Total = \(\frac{2}{9} + \frac{2}{9} = \frac{4}{9}\)

Now QTY B :

\(\frac{20}{30} \times \frac{10}{29}\) = here the denominator will be 29 * 3 that is way big compared to the denominator of QTY A .

Therefore QTY A > QTY B


Why would the denominator remain 30 after selecting a coin?
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
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2 coins are selected with replacement from the bag.

Hence the denominator will be 30

shubhammathur98 wrote:
Why would the denominator remain 30 after selecting a coin?
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
Sir, why does the order matter here?
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
Sir,

The order does not matter. You choose \(1\) coin out of \(30\) and then replace that one coin so the coins available while choosing the second will be \(30\)

Please ask if the doubt remains.

venkyappu19981999 wrote:
Sir, why does the order matter here?
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A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
I understood that point but I did not understand why we are adding 2/9 2 times, I thought its just 2/9
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
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Oh! Yes.

We are selecting two coins and the probability of one fair coin is needed.

Either we get the fair coin on the first chance or we get one fair coin on the second chance. Hence 2/9 for twice.

venkyappu19981999 wrote:
I understood that point but I did not understand why we are adding 2/9 2 times, I thought it's just 2/9
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
1
as we know we need to select 2 coins but we want exactly one fair coin.

then it will be like P(Fair coin) * P(Non fair coin) = 20/30 * 10/30 * 2! = 4/9
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Re: A bag has 30 different coins, of which 20 are fair. 2 coins [#permalink]
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