# Thread: Probability problems due tommorrow

1. ## Probability problems due tommorrow

// Could someone give me a precise and descriptive solution to these problems,,thank you so much!! //

Q1:
if a permutation of the word white is selected at random, find the probability that the permutatation
a.) begins with a consonant b.) ends with a vowel

Q2:
A coded item in a catalog begins with with 3 distinct letters followed by 4 distinct nonzero digits, find the probability of randomly selecting one of these coded items with the first letter a vowel and the last digit even.

Q3:
Two cards are drawn in succession from a deck without replacement. What is the probability that both cards are greater than 2 and less than 8?

2. Originally Posted by jose711
// Could someone give me a precise and descriptive solution to these problems,,thank you so much!! //

Q1:
if a permutation of the word white is selected at random, find the probability that the permutatation
a.) begins with a consonant b.) ends with a vowel
A permutation permits any element of a set to be used not at all or once. The first letter has equal chance of being a W, as it does H, I, T or E. This is a uniform distribution, so to answer your questions:

a) 60%
b) 40%

Note that these numbers should add up to 1, or 100%, because the only options are that the word begin with a consonant or a vowel.

Originally Posted by jose711
Q2:
A coded item in a catalog begins with with 3 distinct letters followed by 4 distinct nonzero digits, find the probability of randomly selecting one of these coded items with the first letter a vowel and the last digit even.
You have no knowledge of what the 2nd and 3rd letters are, so you can simply deduce that the chance of the first letter being a vowel is $\frac{5}{26}$ or $\frac{6}{26}$ if you include y as a vowel. Likewise, the same logic applies to the numbers. The set of digits is 1, 2, 3, 4, 5, 6, 7, 8 and 9. There are four even digits and five odd digits. The probability of picking an even digit is $\frac{4}{9}$.

Therefore, your total probability of randomly selecting a coded item with the first letter a vowel and the last digit even is the product of these two separate probabilities.

Originally Posted by jose711
Q3:
Two cards are drawn in succession from a deck without replacement. What is the probability that both cards are greater than 2 and less than 8?
First, find the probability of the first card meeting the criteria. Greater than 2 and less than 8 implies a 3, 4, 5, 6 or 7. There are five of these in a suit and four suits in a deck making 20 possibilities out of 52. With the first card meeting that criteria, the second card's set of possibilities decreases to a possible 19 out of 51.

The probability that both cards are greater than 2 and less than 8 is the product of these two probabilities.

3. Ill give you a start ...

1. There are 5x4x3x2x1 different permutations of "white".

For each letter, there are 4x3x2x1 permutations where that letter is at the front, or 1/5 of the number of all the permutations.

So there is a 1 in 5 chance that a particular permutation starts with a particular letter.

There are 3 consonants in "white", so there is a 3 in 5 chance that the first letter will be either w, h or t.

The same argument can be used to show that there is a 2 in 5 chance that the random perm ends in a vowel.

Q2. Assuming the letters and numbers in the catalogue name are completely random, there is a 5 in 26 chance that the first letter is a vowel.

As precisely half the digits are even, there is a 1 in 2 chance that the last digit is even.

the probability that both of these are true is 5/26 times 1/2 (as the probs are independent).

Time for my tea so I'll have to stop there.