Please answer the following questions and show and explain how you got your answ
ID: 3303121 • Letter: P
Question
Please answer the following questions and show and explain how you got your answers:
1. Is it more probable to obtain at least one 6 in four throws of a fair die than it is to obtain at least one pair of 6's when two fair dice are thrown simultaneously 24 times?
2. William Stewart's four children come down for breakfeast at random times in the morning. What is the probability that they appear in order from the oldest to the youngest. Note: William Stewart does not have twins( nor triplets, nor quadruplets!)
3. The Computer Science Department at North Carolina State University wishes to market a new software product. The name of this package must consist of three letter's only.
(a.) How many possibilities does the department have to choose from?
(b.) How many possibilities are there if exactly one vowel is included?
(c.) How many possibilities are there if all three characters must be different and exactly one must be a vowel?
Explanation / Answer
1. RECIPE
There are 5 letters out of which one is repeated twice.
Total number of arrangements = 6! / 2! = 720 / 2 = 360.
BOOK
There are 3 letters out of which one is repeated twice.
Total number of arrangements = 4! / 2! = 24 / 2 = 12.
COOKBOOK
There are 4 letters out of which one letter (O) is repeated 4 times and one K is repeated twice.
Total number of arrangements = 8! / (4! 2!) = 40320 / (24*2) = 40320 / 48 = 840.
2. The probability that a 6 DOES NOT appear on a throw of dice = 5/6
=> Probability that a 6 does not appear on 4 throws of dice = (5/6)4
=> Probability that atleast one 6 appears on 4 throws of dice = 1 - (5/6)4
= 1 - 0.48225
= 0.5177
Probability that a pair of 6s appears when two dice are simultaneously thrown = 1/36
=> Probability that a pair of 6s does not appear when two dice are simultaneously thrown = 35/36
=> Probability that no pair of 6s appears when two dice are simultaneously thrown 24 times = (35/36)24
=> Probability that atleast one pair of 6s appears when two dice are simultaneously thrown 24 times = 1 - (35/36)24
= 1 - 0.5086
= 0.4914
Thus the probability of obtaining atleast one 6 in four throws of a dice is greater than the probability of obtaining at least one pair of 6s when two dice are thrown simultaneously 24 times.
3. There are four children and so there are 4! = 24 orders for the children to come down.
Only one among them is the order from oldest to youngest.
The probability that they appear in order from the oldest to the youngest
= 1 / 24
= 0.0417