if three mangoes are taken one after the other from a basket which contains
10 ripe and 4 unripe mangoes.
(a) What is the probability that you will get 3 ripe mangoes?
The probability that you get a ripe mango first is 10 ripe mangos out of 14
mangos. That's 10 out of 14 or 10/14 which reduces to 5/7.
The probability that you get a ripe mango second is 9 ripe mangos out of 13
mangos. That's 9 out of 13 or 9/13 which doesn't reduce.
The probability that you get a ripe mango third is 8 ripe mangos out of 12
mangos. That's 8 out of 12 or 8/12 which reduces to 2/3.
To succeed you must get a ripe one first AND and ripe one second AND a ripe
one third. "AND" indicates that we MULTIPLY. So the PRODUCT of those three
probabilities is (5/7)(9/13)(2/3) = 30/91.
That's the answer.
(b) What is the probability that you will not get any ripe mangoes?
That means you get 3 unripe mangos.
The probability that you get an unripe mango first is 4 unripe mangos out of
14 mangos. That's 4 out of 14 or 4/14 which reduces to 2/7.
The probability that you get an unripe mango second is 3 unripe mangos out
of 13 mangos. That's 3 out of 13 or 3/13 which doesn't reduce.
The probability that you get an unripe mango third is 2 unripe mangos out of
12 mangos. That's 2 out of 12 or 2/12 which reduces to 1/6.
To succeed you must get an unripe one first, AND and unripe one second, AND
an unripe one third. "AND" indicates that we MULTIPLY. So the PRODUCT of
those three probabilities is (5/7)(9/13)(2/3) = 30/91.
The PRODUCT of those three probabilities is (2/7)(3/13)(1/6) = 1/91. That's
the answer.
(c) What is the probability that you will get (exactly) 1 ripe mango?
Case 1. You get a ripe mango first, and an unripe mango second and third.
The probability that you get a ripe mango first is 10 ripe mangos out of 14
mangos. That's 10 out of 14 or 10/14 which reduces to 5/7.
The probability that you get an unripe mango second is 4 unripe mangos out
of 13 mangos. That's 4 out of 13 or 4/13 which doesn't reduce.
The probability that you get an unripe mango third is 3 unripe mangos out of
12 mangos. That's 3 out of 12 or 3/12 which reduces to 1/4.
To succeed with case 1, you must get a ripe one first AND and an unripe one
second AND an unripe one third. "AND" indicates that we MULTIPLY. So the
PRODUCT of those three probabilities is (5/7)(4/13)(1/4) = 5/91. That's the
answer for case 1.
Case 2. You get a ripe mango second, and an unripe mango first and third.
That's the same probability as Case 1, or 5/91.
Case 3. You get a ripe mango third, and an unripe mango first and second.
That's the same probability as Case 1, or 5/91.
To succeed you must have case 1 OR case 2 OR case 3. "OR" indicates that we
ADD. So the SUM of those three probabilities is 5/91 + 5/91 + 5/91 = 15/91,
which does not reduce. That's the answer.
Edwin