Question

How many three-letter "words" can be made from 6 letters "FGHIJK" if repetition of letters

(a) is allowed?

An access code consists of 1 letter of the alphabet followed by 6 digits. (Digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.) How many different access codes are possible?

A jar contains 9 red marbles, numbered 1 to 9, and 12 blue marbles numbered 1 to 12.

a) A marble is chosen at random. If you're told the marble is blue, what is the probability that it has the number 2 on it?

   

b) The first marble is replaced, and another marble is chosen at random. If you're told the marble has the number 1 on it, what is the probability the marble is blue?

The results for a test given by a certain instructor not in the School of Sciences by grade and gender follow.

	A	B	C	Total
Male	10	5	7	22
Female	2	14	16	32
Total	12	19	23	54

Find the probability that a randomly selected student was female given that the student earned a grade of 'B'.

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	4	14	16	34
Female	19	6	3	28
Total	23	20	19	62

If one student is chosen at random,

Find the probability that the student got a 'B' GIVEN they are male.

Answer 1

1) If repetition is allowed then for each place in 3 letter word we have 6 choices.

So, 6*6*6 = 216 words can be made.

2) Given the password contains total 7 places one with letter and other 6 with numbers.

So for first place we have 26 choices of letters.

Now from second place onwards we have to choose numbers from 0-9 that is we have in all 10 numbers from which we have to select. Here there is no mention that numbers can not be repeated. So considering numbers repetition we have 10 choices for each 6 places of numbers.

So, 26*10*10*10*10*10*10 = 26,000,000 different access codes are possible.

3) Given : 9 red marbles and 12 blue marbles.

a) It is said that the marble is blue. So we have total 12 marbles. Now getting number 2 can happen in only 1 way by selecting that marble.

So, p( marble is blue having number 2) = 1/12

b) Now, it is given that drawn marble has number 1. We have 2 marbles with number 1, red and blue. Selecting blue can happen in 1 way.

So, p(marble has number 1and is blue) = 1/2