Solving probability based questions is easy. You just need to get the concept right and know the right formulae. Alright, don’t give me those ‘amuse-me’ expressions, I realize that answering all the questions in the SAT exam to get high SAT scores requires concept and knowledge about the correct usage. Knowledge along with **rigorous, regular and smart SAT testing** is your key to solving the toughest questions and making it to the best colleges in the country.

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Given below are a few probability questions, detailed solutions for which are also given at the end. Go through the questions and solve them before looking at the solutions. Give your best shot and if you falter, try again!

Four letters are to be placed in four addressed envelopes. If the letters are placed at random into the envelopes, find out the following probabilities.

1. The probability that none of the letters are placed in the right envelopes is:

a) 1/24

b) 3/8

c) 23/24

d) 13/24

e) None of the above

2. The probability that at least three letters are placed in the right envelopes is

a) 9/24

b) 1/24

c) 23/24

d) 13/24

e) 0

3. Probability that exactly one letter is placed in the wrong envelope is

a) 19/24

b) 5/24

c) 13/24

d) 1/24

e) 1

4. Probability that exactly two letters are placed in the wrong envelopes is

a) 1/2

b) 1/4

c) 1/3

d) 1/6

e) 1/5

5. Probability that all the letters are placed in their corresponding envelopes is

a) 1/24

b) 1/12

c) 1/8

d) 1/6

e) 1/5

If you still face problems with probability concepts and methods, then go through the probability explanation post on the site.

**Solutions**

There are four letters and four corresponding addressed envelopes. These four letters can be placed in 4! Ways = n(s) = 24

**1.** Given that none of the letters is placed into corresponding envelopes, means that letters are placed in wrong envelopes.

Now, the letters can be placed in wrong envelopes in 9 ways. Therefore, the required probability is

= 9/24 = 3/8. Hence, answer = (b)

**2. **At least three letters placed in right envelopes means that three or more than three letters are placed in corresponding envelopes.

Now, the probability that three letters are placed in the right envelope (only three letters, not the fourth one) is **0 **because of the simple reason that if three are correctly placed, then the fourth one will automatically fall into place.

The probability that all the four letters are placed correctly is 1/24 because there can be only way in which all the letters are correctly placed.

The probability that at least letters are correctly placed = 0 + 1/24 = 1/24. Hence, the answer is (b)

**3. **The probability that only one letter is placed in a wrong envelope is without doubt 0. Need I explain the reason? By the way, the answer is **(e)**

**4. **Exactly two letters are placed in wrong envelopes and we have to find out the probability for that.

Two letters can be selected from 4 letters in ⁴C₂** **ways. This is equal to 6 ways. Now these two can be arranged in a wrong fashion in only 1 way.

Therefore, the probability is 6/24 = ¼. Hence, the answer is **(b)**

**5. **We have already solved this question and the answer is (a) that is 1.

Actual SAT does not have this kind of complicated probability questions. By the way, the answer to #1 is correct or 3/8.

Your answer to #1 is incorrect. As you place the envelopes, the probability of placing the first letter in the wrong envelope is 3/4. IF you placed that one in the wrong envelope, then the probability of placing the second one in the wrong envelope is 2/3. IF you placed that one in the wrong envelope, then the probability of placing the third one in the wrong envelope is

1/2. IF you placed that one in the wrong envelope, then the probability of placing the last one in the wrong envelope is 100%. 3/4 times 2/3 times 1/2 equals 1/4.

So it must be that there are only 6 ways that they can all be wrong, not 9.