10. What is the probability of these events

raja.ir02

raja.ir02

Answered question

2022-01-31

10. What is the probability of these events whenwe randomly select a permutation of the 26 lowercase letters of the English alphabet?

a) The first 13 letters of the permutation are in alphabetical order.

b) a is the first letter of the permutation and z is the last letter.

c) a and z are next to each other in the permutation.

d) a and b are not next to each other in the permutation.

e) a and z are separated by at least 23 letters in the permutation.

f ) z precedes both a and b in the permutation.

Answer & Explanation

alenahelenash

alenahelenash

Expert2022-03-09Added 556 answers

a) The first 13 letters of the premutatuion are in alphabetical order.

probability =263126=1326

d) a and b are not next to each other in the permutation.

We need to se;ect 26 out of 26 letters and we need to determine the number of permutations.

P(26,26)=26!(2626)!=26!0!=26!

We could find that 225! of the 26! permutations will have a and b next to each other and thus 26!225! of the 26! permutations will have a and b not to next to each other.

The probability is the number of favorable outcomes divided by the number of possible outcomes.

P (a and b not next to each other)

=# of favorable outcomes# of possible outcomes

=26!225!26!=312,269,041,039,943,663,616,000,000403,291,461,126,605,635,584,000,000=12130.9231

e) The number of possible permutation of 26 letters if 26!. First find the number of ways to place a and z when any one of aand z is placed at first place in 2 ways then another one of a and z can be places in 2 ways at 25th palce or 26th place.

So number of ways to place a and z is 4 ways. 

Now, when any one of a and z is placed at second place in 2 ways then another one of a and z can be places in 1 ways at 26th place.

So, number of ways to place a and z is 2 ways.

So, total number of ways to place a and z is

4+2=6

The remaining 24 places can be arranged in 24! ways.

So, total number of ways to select a permutation of the 26 letters of the English alphabet in such a way that a and z are separate by at least 23 letters in permutation is 6(24!).

So, the probability is

6(24!)26!=62625

=6650

=0.00923076923

 

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