Question

2/3 Points] DETAILS PREVIOUS ANSWERS WANEFMAC7 8.4.004.NVA MY NOTES PRACTICE ANOTHER Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing four red marbles, three green ones, four white ones, and one purple one. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT (See Example 1.] She has at least one green one. 114 X Need Help? Watch It Talk to Tutor -/3 Points] DETAILS WANEFMAC78.4.006.NVA MY NOTES PRACTICE ANOTHER Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, four green ones, three white ones, and three purple ones. She grabs five of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT (See Example 1.] She has two green ones and one of each of the other colors. Need Help? Watch It Talk to a Tutor 13 Points) DETAILS WANEFMAC7 8.4.010.NVA MY NOTES PRACTICE ANOTHER Recall from Example 1 that whenever Suzan sees a bag of marbles, she grabs a handful at random. She has seen a bag containing three red marbles, four green ones, two white ones, and three purple ones. She grabs eight of them. Find the probability of the following event, expressing it as a fraction in lowest terms. HINT [See Example 1.] She does not have all the green ones. C Need Help? Watch It Talk to a Tutor

Answer 1

(1) There are 4 Red, 3 Green, 4 White and 1 Purple in the Bag

Therefore there is totally 12 Marbles in the Bag

We know that

22 PE) =

Where m = Favourable number of events

n = Total number of events

The way of drawing 8 marbles from the 12 marbles is

$n = \binom{12}{8} = 495$.

m = At least one Green Marble

NOTE: In Probability the word "At least" refers to ADDITION.

In probability the word "AND" refers to MULTIPLICATION

Therefore m = 1green and 7 other (or) 2 green and 6 others (or) 3 green and 5 others

Here Others means; Mixture of Red, White and Purple. Those MIxture we have totally 9 marbles

$m = \binom{3}{1}*\binom{9}{7} + \binom{3}{2}*\binom{9}{6} + \binom{3}{3}*\binom{9}{5}$.

Therefore

$P(E) = \frac{486}{495}$

$\therefore \, \, P(E) = 0.9818$

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(2) There are   3 Red, 4 Green, 3 White and 3 Purple marbles in the bag.

Therefore totally there is 13 Marbles.

We know that

22 PE) =

Where m = Favourable number of events

n = Total number of events

The way of drawing 5 marbles from the 13 marbles is

$n = \binom{13}{5} = 1287$

m = 2 green and one of each of the other colour.

m = 2 green from 4 greee and 1 red from 3 red and 1 white from 3 white and 1 purple from 3 purple

NOTE; In Probability the word "AND" refers to MULTIPLICATION

$m = \binom{4}{2}*\binom{3}{1}*\binom{3}{1}*\binom{3}{1}$

Therefore

$P(E) = \frac{162}{1287}$

$\therefore \, \, \, P(E) = 0.1259$

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(3)

There are 3 Red, 4 Green, 2 White and 3 Purple in the bag.

Therefore there is 12 Marbles in the bag.

We know that

22 PE) =

Where m = Favourable number of events

n = Total number of events

The way of drawing 8 marbles from the 12 marbles is

$n = \binom{12}{8} = 495$

m = No Green Marbles

m = No Green and 8 Others.

Here Others mean Mixture of Red, White and Purple Marbles. Therefore the number of mixture marbles is 8 marbles.

m = No Green from 4 green marbles and 8 other from 8 other marbles.

NOTE: In Probability the word "AND" refers to MULTIPLICATION.

$m = \binom{4}{0}*\binom{8}{8}$

Therefore

$P(E) = \frac{1}{495}$

$\therefore \, \, P(E) = 0.0020$