# A bag contains 8 blue and 4 yellow balls. Two balls are taken randomly at a time from the bag. Find the probability that both balls are yellow? Without replacement: With replacement:

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Dec 6, 2017

With replacement = 1:9
Without replacement = 1:11

#### Explanation:

With replacement:
Each time we take out a ball, the number of balls in the bag stays the same.

This means that we will always have 12 balls in the bag, and there will always be 4 yellow, and 8 blue.

Knowing this, every time we have a 1 in 3 chance of pulling a yellow ball.

Since we are trying to pull 2 yellow balls, the probability of doing so is:
$\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$

Without replacement
Each time we take out a ball, the number of balls in the bag decreases by one.

This means that after the first yellow ball is pulled, there will only be 11 balls in the bag, and only 3 yellow balls remain.

The probability of pulling a yellow ball is equal to the number of yellow balls divided by the total number of balls in the bag.

The probability of pulling 2 yellow balls is:
$\frac{4}{12} \cdot \frac{3}{11} = \frac{1}{11}$

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Dec 6, 2017

Without replacement: $P \left(Y Y\right) = \frac{1}{11}$

With replacement: $P \left(Y Y\right) = \frac{1}{9}$

#### Explanation:

"Probability" = ("number of desirable outcomes")/("total number of possible outcomes")

To have BOTH balls yellow means we want to find:

P(Y Y)" larr the first ball is yellow AND the second ball is yellow.

For the first ball: P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")

If the ball is NOT replaced, there is one less yellow ball and one less ball.
For the second ball: P(Y) = 3/11 " "(larr3 " yellow balls")/(larr 11 " balls altogether")

$P \left(Y Y\right) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{11} \text{ } \leftarrow$ (AND implies multiply)

If the first ball IS replaced, the probability stays the same for the second ball being yellow.

For the second ball: P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")

$P \left(Y Y\right) = \frac{4}{12} \times \frac{4}{12} = \frac{16}{144} = \frac{1}{9} \text{ } \leftarrow$ (AND implies multiply)

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Dec 6, 2017

$\frac{1}{11}$ (without replacement) and $\frac{1}{9}$ (with replacement)

#### Explanation:

1) without replacement:

$\left(\frac{4}{12}\right) \left(\frac{3}{11}\right) = \frac{12}{11 \cdot 12} = \frac{1}{11}$

2) with replacement:

${\left(\frac{4}{12}\right)}^{2} = {\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$

Explanation :
$\frac{4}{12}$ because there are $4$ yellow balls out of $4 + 8 = 12$ total balls so the odds of picking a yellow one is $\left(\frac{4}{12}\right) = \frac{1}{3}$
.
If the ball is replaced, the second time we have the same probability $\frac{4}{12}$ of picking a yellow one.

However if the ball is not replaced, there are only $3$ yellow balls left out of a total of $11$ balls, from there $\frac{3}{11}$.

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