Probability Questions & Answers

2) Un agente de bienes raíces tiene ocho llaves maestras para abrir varias casas nuevas. Sólo 1 llave maest abrirá cualesquiera de las casas. Si \( 40 \% \) de estas casas por lo general se dejan abiertas, ¿cuál es probabilidad de que el agente de bienes raíces pueda entrar en una casa específica, si selecciona 3 lla maestras al azar antes de salir de la oficina?

Assume the random variable \( x \) is normally distributed with mean \( \mu=87 \) and standard deviation \( \sigma=4 \). Find the indicated probability. \( P(74<x<79) \) \( P(74<x<79)=\square \) (Round to four decimal places as needed.)

Se eligen al azar dos números diferentes de \( \{1,2,3,4,5\} \) y se Itiplican. ¿Cuál es la probabilidad que el producto sea par? \( \begin{array}{lllll}\text { A) } 0.2 & \text { (B) } 0.4 & \text { (C) } 0.5 & \text { (D) } 0.7 & \text { (E) } 0.8\end{array} \)

Answer the questions below. \( \begin{array}{l}\text { (a) From a collection of } 50 \text { store customers, } 2 \text { are to be chosen to receive a special gift. How many } \\ \text { groups of } 2 \text { customers are possible? } \\ \text { (b) A company that makes crayons is trying to decide which } 5 \text { colors to include in a promotional } \\ \text { mini-box of } 5 \text { crayons. The company can choose the } 5 \text { mini-box colors from its collection of } 70 \\ \text { colors. How many mini-boxes are possible? } \\ \text { a }\end{array} \)

Answer the questions below. \( \begin{array}{l}\text { (a) To log on to a certain computer account, the user must type in a } 3 \text {-letter password. In such a } \\ \text { password, no letter may be repeated, and only the lower case of a letter may be used. How many } \\ \text { such 3-letter passwords are possible? (There are } 26 \text { letters in the alphabet.) } \\ \text { (b) } 50 \text { athletes are running a race. A gold medal is to be given to the winner, a silver medal is to be } \\ \text { given to the second-place finisher, and a bronze medal is to be given to the third-place finisher. } \\ \text { Assume that there are no ties. In how many possible ways can the } 3 \text { medals be distributed? } \\ \square\end{array} \)

Question 8: Probability ( 9 marks) A bag contains 3 red, 5 blue, and 2 green marbles. Two marbles are drawn at randon (a) What is the probability that both marbles are red?

24. Sea una variable aleatoria que sigue una distribución normal de media 3 y desviación típica 0,8 . Halla el valor de a en cada caso: \( \begin{array}{ll}\text { a) } P(X \leq 2 a)=0,5 & \text {; b) } P\left(X>\frac{a}{2}\right)=0,9991\end{array} \) SOL. a) 1,5 ; b) 5

7.1 in Volledige steekproef word opgemaak vanaf twee komplementêr S en T , sodat \( \mathrm{P}\left(\mathrm{S}^{\prime}\right)=0,33 \). 7.1.1 Voltooi die volgende stelling: \( \mathrm{P}(\mathrm{S})+\mathrm{P}(\mathrm{T})=\ldots \) 7.1.2 Skryf neer die waarde van \( \mathrm{P}(\mathrm{T}) \).

If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has \( A \) objects of one type, while the remaining \( B \) objects are, the other type, and if \( n \) objects are sampled without replacement, then the probability of getting \( x \) objects of type \( A \) and \( n-x \) objects of type \( B \) under the hypergeometric distribution given by the following formula. In a lottery game, a bettor selects six numbers from 1 to 51 (without repetition), and a winning six-number combination is later randomly selected. Fin the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use \( A=6, B=45, n=6 \), and \( x=2 \) ) \( P(x)=\frac{A!}{(A-x)!x!} \cdot \frac{B!}{(B-n+x)!(n-x)!} \div \frac{(A+B)!}{(A+B-n)!n!} \) \( P(2)=\square \) (Round to four decimal places as needed)

If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done witho replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has \( A \) objects of one type, while the remaining \( B \) objects the other type, and if \( n \) objects are sampled without replacement, then the probability of getting \( x \) objects of type \( A \) and \( n-x \) objects of type \( B \) under the hypergeometric distribu given by the following formula. In a lottery game, a bettor selects six numbers from 1 to 51 (without repetition), and a winning six-number combination is later randomly selected the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use \( A=6, B=45, n=6 \), and \( x=2 \) ) \( P(x)=\frac{A!}{(A-x)!x!} \cdot \frac{B!}{(B-n+x)!(n-x)!} \div \frac{(A+B)!}{(A+B-n)!n!} \) \( P(2)=\square \) (Round to four decimal places as needed.)