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goldfiish [28.3K]
3 years ago
10

Box A contains 1 black and 3 white marbles, and box B contains 2 black and 4 white marbles. A box is selected at random, then a

marble is drawn at random from the selected box. Given that the marble is black, find the probability that Box A was chosen.
Mathematics
1 answer:
Anna11 [10]3 years ago
3 0

Answer: Probability that Box A was chosen given that black marble is chosen is 0.5.

Step-by-step explanation:

Since we have given that

Number of boxes = 2

In Box A,

Number of black marbles = 1

Number of white marbles = 3

In Box B,

Number of black marbles = 2

Number of white marbles = 4

Since black marble is selected.

So, using Bayes theorem , we get that

P(E_1|B)}=\dfrac{P(E_1).P(B|E_1)}{P(E_1).P(B|E_1)+P(E_2).P(E_2|B)}\\\\P(E_1|B)=\dfrac{0.5\times \dfrac{1}{3}}{0.5\dfrac{1}{3}+0.5\times \dfrac{2}{6}}\\\\P(E_1|B)}=\dfrac{0.167}{0.167+0.167}\\\\P(E_1|B)}=0.5

Hence, probability that Box A was chosen given that black marble is chosen is 0.5.

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<h3>6 inches</h3>

Step-by-step explanation:

If Casey and Stephanie order a rectangular cake of dimension 24in by 18in, to get the side length of the largest piece of square, we will find the greatest common factor of both dimension as shown:

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8 0
3 years ago
a committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer.
miss Akunina [59]

Answer:

\frac{1}{990}

Step-by-step explanation:

<u>The full question:</u>

<em>"A committee has eleven members. there are 3 members that currently serve as the boards chairman, ranking members, and treasurer. each member is equally likely to serve in any of the positions. Three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. What is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current​ positions?"</em>

<em />

<em />

The permutation of choosing 3 members from a group of 11 would be:

P(n,r) = \frac{n!}{(n-r)!}

Where n would be the total [in this case n is 11] & r would be 3

Which is:

P(11,3) = \frac{11!}{(11-3)!}=\frac{11!}{8!}=11*10*9=990

So there are total of 990 possible way and there is ONLY ONE WAY for them to be reassigned. Hence the probability would be:

1/990

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Answer:

Step-by-step explanation:

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6 0
3 years ago
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