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Probability question

2021 · 20 Jul · Shift 1 · Q35
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  5. /2021 · 20 Jul · Shift 1 · Q35

Probability question

2021 · 20 Jul · Shift 1 · Q35

JEE MainMathematicsProbabilityMCQ+4 / −1
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :
  1. A
    166{1 \over {66}}661​
  2. B
    111{1 \over {11}}111​
  3. C
    19{1 \over {9}}91​
  4. D
    211{2 \over {11}}112​
View written solutionFree

Correct answer: B

  1. Count the total number of letters in EXAMINATION

    The word EXAMINATION has 11 letters: E,X,A,M,I,N,A,T,I,O,NE, X, A, M, I, N, A, T, I, O, NE,X,A,M,I,N,A,T,I,O,N

    Repeated letters are:

    • AAA occurs 222 times
    • III occurs 222 times
    • NNN occurs 222 times

  2. Total number of distinct arrangements

    Using all letters, the total number of distinct words is 11!2! 2! 2!\frac{11!}{2!\,2!\,2!}2!2!2!11!​

  3. Favourable arrangements: M at the 4th position

    Fix MMM at the 4th place.

    Now arrange the remaining 10 letters: E,X,A,I,N,A,T,I,O,NE, X, A, I, N, A, T, I, O, NE,X,A,I,N,A,T,I,O,N

    Among these, repetitions are still:

    • AAA occurs 222 times
    • III occurs 222 times
    • NNN occurs 222 times

    So the number of favourable arrangements is 10!2! 2! 2!\frac{10!}{2!\,2!\,2!}2!2!2!10!​

  4. Compute the probability

    P(M at 4th position)=10!2! 2! 2!11!2! 2! 2!P(\text{$M$ at 4th position})= \frac{\dfrac{10!}{2!\,2!\,2!}}{\dfrac{11!}{2!\,2!\,2!}}P(M at 4th position)=2!2!2!11!​2!2!2!10!​​

    Cancelling common factors, P=10!11!=111P=\frac{10!}{11!}=\frac{1}{11}P=11!10!​=111​

  5. Option check

    • A: 166\frac{1}{66}661​
    • B: 111\frac{1}{11}111​ ✅
    • C: 19\frac{1}{9}91​
    • D: 211\frac{2}{11}112​

    Hence, the correct option is B.

  6. Comparison with stored answer

    Stored correct answer: B

    Our derived answer: B

    They match.

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