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20 Jul 2021, 08:44
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An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. The probability that they are of different colours is
A. 2/5
B. 1/2
C. 8/15
D. 7/15
E. 4/15
A. 2/5
B. 1/2
C. 8/15
D. 7/15
E. 4/15
20 Jul 2021, 10:30
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v12345
Ways to pick R and B are 2*4, but it can also be B and R. => 2*4*2=16
Total ways = 6*5=30
P=\(\frac{16}{30}=\frac{8}{15}\)
C
20 Jul 2021, 23:43
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Hey guys
The thing to remember here is that in probability, AND means multiply, OR means add
We are asked the probability that two balls picked out of six be a different color.
Four are black, two are red
The only possibilities here are
RB
BR
In the first case, what is the probability of picking a red ball AND then picking a black ball?
Remember, AND means multiply
The probability of picking a red ball is \(\frac{2}{6}\) as there are two red balls and six total balls
The probability of then picking a black ball is \(\frac{4}{5}\) as there are four black balls and five balls total remaining
Multiply the two to find the probability of getting RB:
\(\frac{2}{6}\) * \(\frac{4}{5}\) = \(\frac{8}{30} = \frac{4}{15}\)
The probability of getting RB is \(\frac{4}{15}\)
The only other way of getting two balls of different color is first getting a black ball and then getting a red ball
Let's calculate the probability for that
P(B) = \(\frac{4}{6}\) and then
P(R) = \(\frac{2}{5 }\)
Hopefully this makes sense why, first there were four black balls and six total balls, after a black ball was chosen there were two red balls and five total balls remaining
Multiply the two probabilities as they both must happen:
\(\frac{4}{6}\) * \(\frac{2}{5 }\) = \(\frac{8}{30} = \frac{4}{15}\)
So the probability of getting RB is \(\frac{4}{15}\) and the probability of getting BR is \(\frac{4}{15}\)
Now if either RB or BR happens, the conditions of the problem are satisfied
How do we calculate the probability of one thing OR another happening?
We add the probabilities of the two things:
\(\frac{4}{15}\) + \(\frac{4}{15}\) = \(\frac{8}{15}\)
The answer is (C)
The thing to remember here is that in probability, AND means multiply, OR means add
We are asked the probability that two balls picked out of six be a different color.
Four are black, two are red
The only possibilities here are
RB
BR
In the first case, what is the probability of picking a red ball AND then picking a black ball?
Remember, AND means multiply
The probability of picking a red ball is \(\frac{2}{6}\) as there are two red balls and six total balls
The probability of then picking a black ball is \(\frac{4}{5}\) as there are four black balls and five balls total remaining
Multiply the two to find the probability of getting RB:
\(\frac{2}{6}\) * \(\frac{4}{5}\) = \(\frac{8}{30} = \frac{4}{15}\)
The probability of getting RB is \(\frac{4}{15}\)
The only other way of getting two balls of different color is first getting a black ball and then getting a red ball
Let's calculate the probability for that
P(B) = \(\frac{4}{6}\) and then
P(R) = \(\frac{2}{5 }\)
Hopefully this makes sense why, first there were four black balls and six total balls, after a black ball was chosen there were two red balls and five total balls remaining
Multiply the two probabilities as they both must happen:
\(\frac{4}{6}\) * \(\frac{2}{5 }\) = \(\frac{8}{30} = \frac{4}{15}\)
So the probability of getting RB is \(\frac{4}{15}\) and the probability of getting BR is \(\frac{4}{15}\)
Now if either RB or BR happens, the conditions of the problem are satisfied
How do we calculate the probability of one thing OR another happening?
We add the probabilities of the two things:
\(\frac{4}{15}\) + \(\frac{4}{15}\) = \(\frac{8}{15}\)
The answer is (C)
