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3 years ago

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A bag contains 3 red, 4 blue, and 6 yellow marbles. One marble is selected at a time, and once a marble is selected, it is not r

eplaced. Find the probability of choosing two yellow marbles. P(yellow, yellow)

1 answer:

astraxan [27]3 years ago

4 0

The probability of choosing two yellow marbles = 5/26

Step-by-step explanation:

Step 1 :

Given

Total number of marbles in the bag = 13 marbles

Number of yellow marbles = 6.

We choose 2 marbles from the bag without replacing and need to determine the probability that both are yellow.

Step 2 :

Probability of finding yellow marble in the first try = Number of yellow marbles / Total number of marbles = $\frac{6}{13}$

Once the yellow marble is selected we have 5 yellow marbles among 12 remaining marbles in the bag.

Probability of finding yellow marble in the second try = Number of yellow marbles / Total number of marbles = $\frac{5}{12}$

Step 3:

P(yellow in first 2 tries) = P(yellow in first try) * P(yellow in second try)

P(yellow in first 2 tries) = $\frac{6}{13} * \frac{5}{12}$ = $\frac{5}{26}$

Step 4:

Answer:

The probability of choosing two yellow marbles = 5/26

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Step-by-step explanation:

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$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

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$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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3 years ago