A four-digit number (with no repetitions) is to be formed from the set of digits {1, 2, 3, 4, 5, 6}. Find the probability that the number is even.
A number is even if its last digit is even. Half of the digits in the set are even, which means that the probability that the number is even is $\frac{1}{2}$.
A four-digit number (with no repetitions) is to be formed from the set of digits {1, 2, 3, 4, 5, 6}. Find the probability that the number is odd.
A number is odd if its last digit is odd. Half of the digits in the set are odd, which means that the probability that the number is odd is $\frac{1}{2}$.
Three-letter ‘words’ are to be made by arranging the letters of the word COMPUTER. What is the probability that the ‘word’ will start with a vowel?
Three of the eight letters in COMPUTER are vowels, therefore the probability that the word will start with a vowel is $\frac{3}{8}$.
Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that the letter H is first.
The probability that the letter H is first is equal to $\frac{1}{5}$, as there are five letters that could possibly be first, and H is one of them.
\[\begin{aligned} \text{First we find out the number of possible choices of three letters from the word HEART.} \\ \text{This is equal to } \binom{5}{3}\text{, or 10.} \\ \text{Now we must find out the number of possible choices that include the letter H.} \\ \text{To do this, we must choose two other letters.} \\ \text{The number of these selections is equal to } \binom{4}{2}\text{, or 6.} \\ \text{Hence, the total probability of the letter H being chosen is }\frac{6}{10}\text{, or }\frac{3}{5}. \end{aligned}\]Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that the letter H is chosen.
\[\begin{aligned} \text{First we find out the number of possible choices of three letters from the word HEART.} \\ \text{This is equal to } \binom{5}{3}\text{, or 10.} \\ \text{Now we must find out the number of possible choices that include both vowels.} \\ \text{To do this, we must choose one other letter, so the number of choices is 3.} \\ \text{Hence, the total probability of the both vowels being chosen is }\frac{3}{10}\text{.} \end{aligned}\]Three letters are chosen at random from the word HEART and arranged in a row. Find the probability that both vowels are chosen.
\[\begin{aligned} \text{The number of ways that 6 people can be arranged is equal to }6!\text{, or 720.} \\ \text{The number of ways that 4 people can be arranged is equal to }4!\text{, or 24.} \\ \text{We then multiply this by the number of ways women can be placed at the end, or }^3P_2\text{, which is 6.} \\ \text{Hence, the final result is equal to }\frac{24 \times 6}{720}\text{, or }\frac{1}{5}. \end{aligned}\]Three men and three women are to be randomly seated in a row. Find the probability that both the end places will be filled by women.
\[\begin{aligned} \text{The number of possible 7-person teams selected from 13 people is equal to }\binom{13}{7}\text{, or 1716.} \\ \text{Now we must calculate the number of teams that contain 4, 5, or 6 men.} \\ \text{This is equal to } \binom{6}{4}\binom{7}{3} +\binom{6}{5}\binom{7}{2} +\binom{6}{6}\binom{7}{1} , \text{which is equal to 658}.\\ \text{Therefore, the total probability that the team contains more men than women is }\frac{658}{1716}\text{, or }\frac{329}{858}. \end{aligned}\]A netball team of seven players is to be chosen from six men and seven women. Find the probability that the selected team contains more men than women.
Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that the sandwich contains ham.
Bill has an equally likely chance to both pick or not pick ham. Therefore, the chance that Bill picks ham is equal to $\frac{1}{2}$.
Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that the sandwich contains three ingredients.
The total number of possible choices of ingredients that Bill can pick is equal to $2^8$, or 256. The number of ways that he can pick three ingredients is equal to $\binom{8}{3}$, or 56. Therefore the total probability of Bill picking three ingredients is equal to $\frac{56}{256}$, or $\frac{7}{32}$.
Bill is making a sandwich. He may choose any combination of the following: lettuce, tomato, carrot, cheese, cucumber, beetroot, onion, ham. Find the probability that the sandwich contains at least three ingredients.
The total number of possible choices of ingredients that Bill can pick is equal to $2^8$, or 256. The number of ways that he can pick at least three ingredients is equal to $\binom{8}{3}+\binom{8}{4}+\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}$, or 219. Therefore the total probability of Bill picking three ingredients is equal to $\frac{219}{256}$.
A bag contains five white, six red and seven blue balls. If three balls are selected at random, without replacement, find the probability they are all red.
The total number of ways to pick three balls from 18 is equal to $\binom{18}{3}$, or 816. The total number of ways to pick three balls from 6 is equal to $\binom{6}{3}$, or 20. Therefore, the total probability of picking three red balls from the bag is equal to $\frac{20}{816}$, or $\frac{5}{204}$.
A bag contains five white, six red and seven blue balls. If three balls are selected at random, without replacement, find the probability they are all different colours.
The total number of ways to pick three balls from 18 is equal to $\binom{18}{3}$, or 816. The total number of ways to pick three different coloured balls is one. Therefore, the total probability of picking three different coloured balls from the bag is equal to $\frac{1}{816}$.