Tanya Prepared 4 Different Letters To Be Sent To 4 Different Addresses GMAT Problem Solving

Question: Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

1/24
1/8
1/4
1/3
3/8

“Tanya prepared 4 different letters to be sent to 4 different addresses”- is a topic of the GMAT Quantitative reasoning section of GMAT. GMAT quant determines the analytical and mathematical skills of the students. The student should select the valid option as an answer by justifying them with proper explanations depending on mathematical calculations. The students must have a basic idea of qualitative in order to decode GMAT Problem Solving questions. The GMAT Quant topic in the problem-solving part incorporates mathematical problems that can be solved by mathematical calculations. The candidates can polish up their understanding by practising questions from the “GMAT Official Guide Quantitative Review”.

Solution and Explanation:

Approach Solution 1:

The problem statement suggests that

Given:

Tanya prepared 4 different letters to be sent to 4 different addresses
4 letters are randomly put into the 4 envelopes

Find out:

The probability that only 1 letter will be in the envelope with its correct address.

Let’s assume that Tanya has just one letter and two envelopes.

Let one envelope is correctly addressed and the other be addressed wrongly.

To solve this question, we have to analyse the number of ways the letter can be put into the envelope.
Since there are two envelopes, Tanya has two ways of placing the letter in the envelopes.
Therefore, there exists only one way in which the letter can go to the correct addressed envelope.
Hence, the probability of putting the letter in the correct envelope is ½
As per the question, let’s say Tanya has one letter but four envelopes.
Therefore, only one envelope notes the address correctly as per the letter. The rest of the three letters are thereby wrongly addressed.
Hence, the probability of putting the letter in the correct envelope is ¼
If Tanya has 3 incorrect envelopes, then she can put the letter wrongly with a probability of ¾

To solve the problem, we need to break the problems into four events.

Event 1:
As we know that the probability of putting one letter in the correct envelope is ¼
Since one letter is kept in the envelope correctly, Tanya is therefore left with three letters and three envelopes.

Event 2:
There are 3 envelopes left and out of which, one envelope cites the correct address for letter number 2. The other two envelopes have the wrong address and the letter number 2 could be placed in either of these two envelopes to meet the condition.

Thus, the probability of this event is ⅔

As per the calculation, the probability of putting the letter number 1 in the correct envelope = ¼
As per the calculation, the probability of putting the letter number 2 in the incorrect envelope =⅔

Event 3:

As per the question, let’s consider letter number 3 to be put into the wrong envelope.

There are 2 envelopes left and therefore the probability of putting this letter number 3 in the wrong envelope ( or equally in the correct envelope) is ½.

Event 4:

Let’s take into consideration letter number 4. This letter has left with one way of putting in.
Therefore, the probability of putting the letter number 4 into the incorrect envelope is 1.
Hence, we can say that the grand event is the combination of all these four events. The four events happen with a probability of their own.
Therefore the total probability of the grand event = 1/4 * 2/3 * 1/2 * 1/1 = 1/12
Since all of these events seem to be mutually exclusive. The events do not occur all at the same time whereas only one letter at a single time.
Therefore, the total probability will be the sum of all the individual probabilities = 1/12 + 1/12 + 1/12 + 1/12 = 1/3

Hence, D is the correct answer.

Approach Solution 2:

The problem statement suggests that

Given:

Tanya organised 4 different letters to be dispatched to 4 different addresses
4 letters are randomly put into the 4 envelopes

Find out:

The probability that only 1 letter will be in the envelope with its correct address.

The total number of ways of allocating 4 letters into 4 envelopes = 4! = 24.
The probability of putting only one letter in the correct envelope = number of envelopes * the number of ways likely to put 3 letters wrongly in the envelopes. = 4*2
Let the envelopes be ABCD.
Then the letters will be either ACDB or ADBC (since A is the correct envelope and the other three envelopes have 2 possible wrong arrangements).
Therefore, the total number of ways since we have four letters= 4*2 = 8
P(C= 1) = 8/24 = 1/3

Hence, D is the correct option.

Approach Solution 3:

The problem statement suggests that

Given:

Tanya organised 4 different letters to be sent to 4 different addresses
4 letters are haphazardly placed into the 4 envelopes

Find out:

The probability that only 1 letter will be in the envelope with its correct address.

The probability of putting the first letter in the correct envelope = ¼
The probability of putting the second letter in the wrong envelope = ⅔
The probability of putting the third letter in the wrong envelope= ½
The probability of putting the fourth letter in the wrong envelope=1
Therefore, ¼ * ⅔ * ½ = 1/12
Since there are 4 letters and the probability of putting one letter in the correct envelope

= 1/12 * 4 = 4/12 = 1/3

Hence, D is the correct answer.

Suggested GMAT Quant Questions

SHARE ON FACEBOOK SHARE ON TWITTER

Tanya Prepared 4 Different Letters To Be Sent To 4 Different Addresses GMAT Problem Solving

Fees Structure

Comments

SUBSCRIBE TO OUR NEWS LETTER