A bag contains 10 marbles,7 of which are black and 3 are red.three marbles are drawn at one after the other.find the probability
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Answer: n = 10.
Step-by-step explanation:
In the bag, we have n counters.
4 of the counters are red.
the rest are blue, then we have:
(n - 4) blue counters.
Now, the probability that Ross takes a blue counter from the bag is equal to the quotient between the number of blue counters (n - 4) and the total number of counters, n
Then the probability is:
p1 = (n - 4)/n
Now he draws another, and it must be blue again, then we can calculate the probability in the same way as above, but he already take a blue counter, so the number of blue counters is (n - 5) and the total number of counters is (n - 1)
The probability of this event is:
p2 = (n - 5)/(n - 1)
The joint probability (the probability that Ross takes two blue counters) is equal to the product of the individual probabilities, and we know that this is equal to 1/3, then we have the equation:
1/3 = ( (n - 4)/n)*((n - 5)/(n - 1))
Now let's solve this for n.
n*(n - 1)/3 = (n - 4)*(n - 5)
(n^2 - n)/3 = n^2 - 4*n - 5*n + 20
n^2 - n = 3*(n^2 - 9*n + 20)
n^2 - n = 3*n^2 - 27*n + 60
0 = (3*n^2 - n^2) - 27*n + n + 60
0 = 2*n^2 - 26*n + 60
The two solutions of this equation can be found with Bhaskara's equation:
Then the two solutions are:
n = (26 - 14)/4 = 3
This is not an option, because we know for sure that we have 4 red counters, then this option can be discarded.
The other solution is:
n = (26 + 14)/4 = 40/4 = 10
Then we have n = 10, 10 counters in total.