Five tiles which are marked individually with the letters A to E are

Question

Five tiles which are marked individually with the letters A to E are placed into a bag, and randomly drawn out one by one without being replaced. What is the probability that the tile with the letter B is pulled out immediately after the tile with the letter A?

A. 1/5
B. 2/5
C. 1/10
D. 1/30
E. 1/120
​​

Question -
I solved it as, group "AB" as one, and the other 4 letters individually. So, now we have 4! waves of desired outcomes. Total number of outcomes in 5!. Answer is 1/5. Is my logic correct?

Answer is 1/5 (A)

Bunuel · Accepted Answer

So, there are 5 letters in the bag: A, B, C, D, and E. We are interested in the following outcomes:

AB XXX : 1/5*1/4 = 1/20;

X AB XX : 3/5*1/4*1/3 = 1/20;

XX AB X : 3/5*2/4*1/3*1/2 = 1/20;

XXX AB  : 3/5*2/4*1/3*1/2*1 = 1/20;

The sum of the above is 1/20 + 1/20 + 1/20 + 1/20 = 4/20 = 1/5.

Answer: A.

OR:  P = Favorable/Total

Number of favorable outcomes:  consider A and B as one unit: {AB}. Four units {AB}, {C}, {D}, and {E} can be arranged in 4! ways. In all of these outcomes B will be immediately after A:
 {AB}{C}{D}{E};
{AB}{C}{E}{D};
{AB}{D}{C}{E};
{AB}{D}{E}{C};
{AB}{E}{C}{D};
{AB}{E}{D}{C};
...
{C}{E}{D}{AB};

Total number of outcomes:  five letters A, B, C, D, and E can be arranged in 5! ways.

P = Favorable/Total = 4!/5! = 1/5.

Answer: A.

Kinshook · Answer

Given: Five tiles which are marked individually with the letters A to E are placed into a bag, and randomly drawn out one by one without being replaced.

Asked: What is the probability that the tile with the letter B is pulled out immediately after the tile with the letter A?

Total ways = 5!
Favorable ways = 4! after considering AB as one single letter

Probability that the tile with the letter B is pulled out immediately after the tile with the letter A = 4!/5! = 1/5

IMO A

Regor60 · Answer

Treating A and B as one unit, there are 4 tiles to be arranged:

4!

There are 5! ways to arrange the tiles

Probability: 4!/5! = 1/5

Posted from my mobile device

Kushchokhani · Answer

Kaur92 The approach as mentioned in Spoiler is fine

Mislead · Answer

Bunuel ,  IanStewart  and  EXPERT  ,

For this problem can we not take probability of selecting A which will be 1/5 and then multiply it with the conditional probability of having B which is 1 (100% probability), that will give us 1/5*1 = 1/5

is this method correct or i need to revisit my probability chapters ?

Krunaal · Answer

Imo, this reasoning is only valid if A is the first student to be picked. In the given question A can be picked at any position (except last), and the tiles aren't replaced, so the probability of picking A will change relative to the position it is picked.

Regor60 · Answer

The probability of A being picked in any position is 1/5.

The probability of B being picked immediately after A is 1/4 for every position of A EXCEPT for A being picked last.

So the total probability is:

1/5*1/4*4 = 1/5

luisdicampo · Answer

Deconstructing the Question

We have  5  distinct tiles:  A, B, C, D, E .

They are drawn one by one without replacement, and we want the probability that  B  is drawn immediately after  A .

The key idea is to treat  AB  as one block. Since the order must be exactly  A  then  B , the internal order of the block is fixed.

Step-by-step

The total number of possible orders of the  5  tiles is:

5! = 120

Now count the favorable outcomes.

If  A  and  B  must appear as the block  AB , then the objects are:

AB, C, D, E

That gives us  4  objects to arrange.

The number of such arrangements is:

4! = 24

We do not multiply by  2 , because  BA  is not allowed. The order must be exactly  AB .

So the probability is:

\frac{24}{120} = \frac{1}{5}

Answer: A)  \frac{1}{5}

Five tiles which are marked individually with the letters A to E are

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Five tiles which are marked individually with the letters A to E are

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