Suppose that 1 fruit each is drawn from 3 baskets.
The first basket contains 3 oranges, 2 mangoes and 1 apple, the second one contains 2 oranges, 2 mangoes and 2 apples and ,finally, the third one contains 1 orange, 4 mangoes and 3 apples.
What is the probability that all 3 drawn fruits are different?
Probability that 3 drawn fruits are different
combinationscombinatoricsprobability
Related Solutions
You can give any combo of $0-5$ apples, $0-10$ mangoes to person $A$
Thus $6\times11 = 66$ ways (the balance needed to make $15$ will be oranges)
Hint: Let $f(n)$ be the number of ways to eat $n$ pieces of fruit ending with an apple. Because we get stuck on apples, let $g(n)$ be the number of ways to eat $n$ pieces of fruit ending with a banana. There are also $g(n)$ ways to eat $n$ pieces of fruit and ending with an orange. Let $h(n)$ be the number of ways to eat $n$ pieces of fruit ending with strawberries. You should be able to write coupled recurrences for $g(n),h(n)$. Then the total number of ways of eating $n$ pieces of fruit is $f(n)+2g(n)+h(n)$
An apple can go after anything, so $f(n)=f(n-1)+2g(n-1)+h(n)$. A banana can go after an orange or a strawberry, so $g(n)=g(n-1)+h(n-1)$. A strawberry can go after anything but an apple, so $h(n)=2g(n-1)+h(n-1)$
A spreadsheet with the result is below
Best Answer
The six different combinations are
$P(O,M,A) = \frac{3}{6}\frac{2}{6}\frac{3}{8}$
$P(O,A,M) = \frac{3}{6}\frac{2}{6}\frac{4}{8}$
$P(M,O,A) = \frac{2}{6}\frac{2}{6}\frac{3}{8}$
$P(M,A,O) = \frac{3}{6}\frac{2}{6}\frac{1}{8}$
$P(A,M,O) = \frac{1}{6}\frac{2}{6}\frac{1}{8}$
$P(A,O,M) = \frac{1}{6}\frac{2}{6}\frac{4}{8}$
Sum them all you get the probability that all 3 balls are different
Goodluck