A bag contains 4 red and 4 black balls When 3 balls are chosen at random, without replacement, the probability of getting 2 red balls and 1 black ball is
[tex] \frac{n}{m} [/tex]
Find
[tex](m + n)[/tex]
[tex] \frac{n}{m} [/tex]
Find
[tex](m + n)[/tex]
Jawaban:
8
Penjelasan dengan langkah-langkah:
To solve this problem, we can use combinations and the formula for probability.
There are a total of 8 balls in the bag, so the number of possible ways to choose 3 balls is:
C(8,3) = 56
The number of ways to choose 2 red balls and 1 black ball can be found by multiplying:
C(4,2) * C(4,1) = 6 * 4 = 24
So the probability of getting 2 red balls and 1 black ball is:
24/56 = 3/7
Therefore, n/m = 3/7.
To find (m + n), we need to determine the values of n and m separately.
Let x be the value of n. Then, the value of m is 8 - x (since there are a total of 8 balls in the bag).
We know that n/m = 3/7, so:
x / (8 - x) = 3/7
Multiplying both sides by (8 - x), we get:
x = 24/5
So, the value of x (which is equal to n) is 24/5, and the value of m is:
8 - x = 8 - 24/5 = 16/5
Therefore, (m + n) = (16/5 + 24/5) = 8.
Hence, the value of (m + n) = 8.