Question

a local bakery has determined a probability distribution for the number of cheesecakes that given day.
x = #sold 0 5 10 15 20
probability 0.1 0.12 0.38 ...? 0.07
what is the probability of selling 15 cheesecakes in a given day?
what is the probability of selling at least 10 cheesecakes?
what is the probability of selling 5 or 15 cheesecakes?
what is the probability of selling 25 cheesecakes?
give the expected number of cheesecakes sold in a day using the discrete probability distribution?
what is the probability of selling at most 10 cheesecakes?

Explanation:

Step1: Find probability of selling 15 cheesecakes

The sum of all probabilities in a probability - distribution is 1. Let the probability of selling 15 cheesecakes be $P(X = 15)$. Then $0.1+0.12 + 0.38+P(X = 15)+0.07=1$.
$P(X = 15)=1-(0.1 + 0.12+0.38 + 0.07)=1 - 0.67 = 0.33$.

Step2: Find probability of selling at least 10 cheesecakes

$P(X\geq10)=P(X = 10)+P(X = 15)+P(X = 20)$.
$P(X\geq10)=0.38 + 0.33+0.07=0.78$.

Step3: Find probability of selling 5 or 15 cheesecakes

Since the events are mutually - exclusive, $P(X = 5\ or\ X = 15)=P(X = 5)+P(X = 15)$.
$P(X = 5\ or\ X = 15)=0.12+0.33 = 0.45$.

Step4: Find probability of selling 25 cheesecakes

Since 25 is not in the given distribution, $P(X = 25)=0$.

Step5: Find expected number of cheesecakes sold

The formula for the expected value $E(X)$ of a discrete random variable is $E(X)=\sum_{i}x_{i}P(X = x_{i})$.
$E(X)=0\times0.1+5\times0.12 + 10\times0.38+15\times0.33+20\times0.07$
$E(X)=0 + 0.6+3.8+4.95 + 1.4$
$E(X)=10.75$.

Step6: Find probability of selling at most 10 cheesecakes

$P(X\leq10)=P(X = 0)+P(X = 5)+P(X = 10)$.
$P(X\leq10)=0.1+0.12+0.38 = 0.6$.

Answer:

Probability of selling 15 cheesecakes: 0.33
Probability of selling at least 10 cheesecakes: 0.78
Probability of selling 5 or 15 cheesecakes: 0.45
Probability of selling 25 cheesecakes: 0
Expected number of cheesecakes sold: 10.75
Probability of selling at most 10 cheesecakes: 0.6