Question relating to probablilties

[adeel]

I have a test tomorrow and there are 2 question I am stuck on

Five patrons check their coats at a theatre. In the confusion at the end of the performance, the attendant, in despair, hands the coats back at random

What is the probability taht 3 or more of the patrons will get their coat back.

The hint given is: can exactly 4 people receive their correct coat? I am pretty sure they can't because to do that, the fifth person would get the right coat back as well.

second one:

You are doing some wordprocessing, and you accidentally type 3 letters if the alphabet without noticing which ones you typed. What is the probability that the letters will be in alphabetical order from left to right?

So can anyone help? I knoe what the asnwers are, i just dotn know how to get them (the answer is 11/120 and 1/6)

somone, please explain!

 

[metacristi]

1.

The hint given is: can exactly 4 people receive their correct coat? I am pretty sure they can't because to do that, the fifth person would get the right coat back as well.

This means that there is a single case when 4 or more of the patrons will get their coat back.

In how many ways can the attendant hand back the coats?

N=permutations of 5=5!=120

Next it must be calculated the number of possible cases when the attendant hands exactly 3 correct coats back.

N[3]=C₅³=5!/(3!)*(5-3)!=10

The probability that 3 or more of the patrons will get their coat back is:

P=(N[3]+1)/120=11/120


2.I assume that the three letters are distinct.Whatever letters are typed we can form 3!=6 distinct possibilities with them,function of their position (on the first,second or third place).

For example if the three letters were a,d,e we have the possibilities:

a,d,e
a,e,d
d,e,a
d,a,e
e,a,d
e,d,a

In how many cases are they in the alphabetical order?In exactly one case.

Hence the seeked probability is:P=1/6.

The same is valid no matter the letters written,assuming they are distinct.

 

[tacman]

For the first question, the probability of getting 3 or more patrons to receive their correct coat back can be calculated using the binomial distribution formula. The formula is P(X=x) = nCx * p^x * (1-p)^(n-x), where n is the number of trials (in this case, 5 patrons), x is the number of successes (in this case, 3 or more patrons getting their correct coat back), and p is the probability of success (in this case, 1/5 since there is a 1 in 5 chance of each patron getting their correct coat back).

Using this formula, we can calculate the probability as follows:

P(X=3) = 5C3 * (1/5)^3 * (4/5)^2 = 10 * (1/125) * (16/25) = 0.128

P(X=4) = 5C4 * (1/5)^4 * (4/5)^1 = 5 * (1/625) * (4/5) = 0.0032

P(X=5) = 5C5 * (1/5)^5 * (4/5)^0 = 1 * (1/3125) * 1 = 0.00032

Therefore, the total probability of getting 3 or more patrons to receive their correct coat back is 0.128 + 0.0032 + 0.00032 = 0.13152 or approximately 13.15%.

For the second question, the probability of typing 3 letters in alphabetical order can be calculated by considering all the possible combinations of 3 letters from the alphabet (26 letters) and dividing it by the total number of possible combinations of 3 letters (26*25*24).

So, the total number of possible combinations of 3 letters from the alphabet is 26*25*24 = 15600.

Out of these, there are 6 possible combinations that are in alphabetical order (ABC, BCD, CDE, DEF, EFG, FGH).

Therefore, the probability of typing 3 letters in alphabetical order is 6/15600 = 1/2600 = 1/6.

I hope this helps! Good luck on your test tomorrow! Remember to always show your work and double check your calculations.