Recall that situations that may be modelled by a binomial distribution are those in which there are a number of independent trials of the same experiment and the individual outcomes are either success or failure. We are interested in the probability that there will be some number $k$k successes among the $n$n trials of the experiment.
Census data shows that $30%$30% of the population in a particular country have red hair. Find the probability that more than half of a random sample of $6$6 people have red hair.
Give your answer correct to four decimal places.
A science exam consisted of $48$48 multiple choice questions, each with $4$4 possible options. Ray guessed the answers to all of the questions at random.
Let $X$X be the number of questions Ray gets correct.
Find $E\left(X\right)$E(X).
Find $Var\left(X\right)$Var(X).
Find the standard deviation $SD\left(X\right)$SD(X).
Mohamad is currently applying for graduate job positions. For each application he submits, the probability that it gets short-listed and he gets invited for an interview is $0.02$0.02.
If he applies for $8$8 positions, what is the probability that he will not get a single interview?
Round your answer to two decimal places.
If he applies for $8$8 positions, what is the probability that he will get at least one interview?
Round your answer to two decimal places.
If he applies for $n$n positions, what is the probability that he will get at least one interview?
Hence find the minimum number of applications he will have to submit to ensure that the probability that he gets at least one interview is greater than $0.95$0.95.
Mohamad has been applying for $8$8 positions every week over the last $12$12 weeks. Define the random variable $Y$Y, where $Y$Y is the number of weeks Mohamad had at least one interview.
Using your final answer from part (b), complete the following.
$Y$Y$\sim$~$B$B$\left(\editable{},\editable{}\right)$(,) | ||
$P\left(Y=y\right)$P(Y=y) | $=$= | $\binom{\editable{}}{\editable{}}\times0.15^{\editable{}}\times0.85^{\editable{}}$()×0.15×0.85 |
Calculate the probability that Mohamad had at least one interview per week, for at most $3$3 out of the $12$12 weeks.
Round your answer to two decimal places.
Maria buys a box of $8$8 glasses which is on sale because the box is expected to have around $3$3 broken glasses inside.
Solve for the probability $p$p that a glass is broken.
What is the probability that at most half of the glasses are broken?
Round your answer to two decimal places.
At the store there are $12$12 boxes in the sale, each containing $8$8 glasses. All of the boxes are expected to have about $3$3 broken glasses. Define the random variable $X$X, where $X$X is the number of boxes containing at most half of the glasses broken.
Complete the following.
$P\left(X=x\right)$P(X=x) | $=$= | $\binom{\editable{}}{\editable{}}\times0.86^{\editable{}}\times0.14^{\editable{}}$()×0.86×0.14 |
Hence determine the probability that there are exactly $10$10 boxes on display where at most half of the glasses inside are broken.
Round your answer to two decimal places.