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A jar contains a mixture of 12 black marbles, 10 red marbles, and 18 white marbles, all the same size. If two marbles are drawn from the jar without being replaced, what would the probability be:


of drawing a white, then a black marble?


of drawing two black marbles?


of drawing two white marbles?


of drawing a black, then a red marble?

Please help

Verified Answer :

P( two white) = 18/40 17/39 = 51/260
P( black then red) = 12/40 10/39 = 1/13
P( two black) = 12/40 11/39 = 11/130
P( white then black) = 18/40 12/39 = 9/65

Answer:

1) Probability of drawing a white, then a black marble? = 0.13

2) Probability
of drawing two black marbles =0.08

3) Probability
of drawing two white marbles = 0.19  

4) Probability
of drawing a black, then a red marble = 0.07    

Step-by-step explanation:

Given : A jar contains a mixture of 12 black marbles, 10 red marbles, and 18 white marbles, all the same size. If two marbles are drawn from the jar without being replaced.

Solution :

Total number of marbles in a jar = 12+10+18=40

\(\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}\)

1) To find  what would the probability of drawing a white, then a black marble?

probability of getting a white marble = \(\frac{18}{40}=\frac{9}{20}\)

Without replacement, total number is 40-1=39

probability of getting a black marble = \(\frac{12}{39}=\frac{4}{13}\)

Probability of drawing a white, then a black marble is

\(P=\frac{9}{20}\times \frac{4}{13}= \frac{36}{260}=0.13\)


2) To find  what would the probability of drawing two black marbles?  

probability of getting one black marble = \(\frac{12}{40}=\frac{3}{10}\)

Without replacement, total number is 40-1=39

probability of getting second black marble = \(\frac{11}{39}\)

Probability of drawing two black marbles is

\(P=\frac{3}{10}\times \frac{11}{39}= \frac{33}{390}=0.08\)


3) To find what would the probability of drawing two white marbles?  

probability of getting one white marble = \(\frac{18}{40}=\frac{9}{20}\)

Without replacement, total number is 40-1=39

probability of getting second white marble = \(\frac{17}{39}\)

Probability of drawing two white marbles is

\(P=\frac{9}{20}\times \frac{17}{39}= \frac{153}{780}=0.19\)


4) To find what would the probability
of drawing a black, then a red marble?  

probability of getting a black marble = \(\frac{12}{40}=\frac{3}{10}\)

Without replacement, total number is 40-1=39

probability of getting a red marble = \(\frac{10}{39}\)

Probability of drawing a white, then a black marble is

\(P=\frac{3}{10}\times \frac{10}{39}= \frac{30}{390}=0.07\)