GATE Overflow Test Series | Probability | Test 1 | Question: 8

Question

A bag has $8$ pieces of paper, each with one of the letters of $\text{‘GATE2021’}.$  If the probability that the letters of the papers picked in order without replacement spell $\text{‘GATE2021’}$ is $X,\;10000X = $ _____ (Rounded to two decimal points)

Answer 1

Here, only $2$ is repeated. So, in terms of the favorable permutation, there is only two possible order. For $8$ letters total possibility is $8!$ orders. So, our required probability $ = \dfrac{2}{8!} = \dfrac{1}{5040 \times 4} = 0.0000496$

So, $10000X=0.496$

Comments

total 8 letter possibility is 8!/2!. So, the ans will be 2/(8!/2!). Please do correct me if i am wrong.

---

Sir,

in total $\frac{8!}{2!}= 4\cdot 7!\; possibility \; out\: of\: only\: 2 \: favourable\: case \: \\hence \; 10000X= 10000\cdot \frac{2}{4\cdot 7!}= 0.992$

correct me if I’m wrong.

---

In your total cases, which two are the favorable ones?

---

Sorry sir, only 1 favourable case is possible. my bad.

Answer 2

 

For 8 letters total possibility is 8! 

 favorable permutation =2  (2 is repeated here)

so 

 required probability =2/ 8!

                                 =2/40320

                                 = 1/20160

                                 = 0.0000496

So                   10000X=   10000 x 0.0000496 = 0.496

i hope my answer helps you a lot.