Question #5ac9e

Question

Question #5ac9e

2 Answers

Impact of this question

1127 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-06T18:20:57

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

Explanation:

1) without replacement:

$(\frac{4}{12}) (\frac{3}{11}) = \frac{12}{11 \cdot 12} = \frac{1}{11}$ $(4/12)(3/11) = 12/(11*12) = 1/11$

2) with replacement:

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

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

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

Answer 2 · 2017-12-06T18:59:55

Without replacement: $P (Y Y) = \frac{1}{11}$ $P(YY)=1/11$

With replacement: $P (Y Y) = \frac{1}{9}$ $P(YY) = 1/9$

Explanation:

$Probability = \frac{number of desirable outcomes}{total number of possible outcomes}$ $"Probability" = ("number of desirable outcomes")/("total number of possible outcomes")$

To have BOTH balls yellow means we want to find:

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

For the first ball: $P (Y) = \frac{4}{12} \frac{\leftarrow 4 yellow balls}{\leftarrow 12 balls altogether}$ $P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")$

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

$P (Y Y) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{11} \leftarrow$ $P(YY) = 4/12xx3/11 = 1/11" "larr$ (AND implies multiply)

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

For the second ball: $P (Y) = \frac{4}{12} \frac{\leftarrow 4 yellow balls}{\leftarrow 12 balls altogether}$ $P(Y) = 4/12" "(larr4 " yellow balls")/(larr 12 " balls altogether")$

$P (Y Y) = \frac{4}{12} \times \frac{4}{12} = \frac{16}{144} = \frac{1}{9} \leftarrow$ $P(YY) = 4/12xx4/12 = 16/144 =1/9" "larr$ (AND implies multiply)

Science

Math

Humanities

... and beyond

Question #5ac9e

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question