Question

1. 162 students are receiving awards at an awards banquet. 48 receive math awards, 78 receive english awards, and 14 receive both math and english awards. what is the probability that a randomly chosen student has won an award for english or math?

$p(e \text{ or } m)=p(e)+p(m)-p(e \text{ and } m)=$

1. the probability of rain on saturday or sunday is given as 0.5. the probability of rain on saturday is 0.2 and the probability of rain on sunday is 0.4. what is the probability that it will rain on both saturday and sunday?

1. events c and d are such that $p(c)=0.44, p(d)=0.61$, and $p(c^{c} \text{ and } d)=0.23$. determine the value of $p(c^{c} cup d)$

Explanation:

Question 4

Step1: Calculate the number of students who won English or Math awards

Use the formula \(n(E\cup M)=n(E)+n(M)-n(E\cap M)\). Here, \(n(M) = 48\), \(n(E)=78\), \(n(E\cap M)=14\).
\(n(E\cup M)=48 + 78-14=112\)

Step2: Calculate the probability

The probability \(P(E\cup M)=\frac{n(E\cup M)}{n(\text{total})}\). Since \(n(\text{total}) = 162\) and \(n(E\cup M)=112\).
\(P(E\cup M)=\frac{112}{162}=\frac{56}{81}\approx0.691\)

Step1: Use the addition rule of probability

The addition rule is \(P(S\cup Su)=P(S)+P(Su)-P(S\cap Su)\). We know that \(P(S\cup Su) = 0.5\), \(P(S)=0.2\), \(P(Su)=0.4\)

Step2: Solve for \(P(S\cap Su)\)

Rearrange the formula \(P(S\cap Su)=P(S)+P(Su)-P(S\cup Su)\)
Substitute the values: \(P(S\cap Su)=0.2 + 0.4-0.5=0.1\)

Step1: Find \(P(C^{C})\)

Since \(P(C)+P(C^{C}) = 1\), then \(P(C^{C})=1 - P(C)=1 - 0.44 = 0.56\)

Step2: Use the formula \(P(C^{C}\cup D)=P(C^{C})+P(D)-P(C^{C}\cap D)\)

We know that \(P(C^{C}\cap D) = 0.23\) and \(P(C^{C})=0.56\), \(P(D)=0.61\)
Substitute into the formula: \(P(C^{C}\cup D)=0.56+0.61 - 0.23\)
\(P(C^{C}\cup D)=0.94\)

Answer:

The probability that a randomly - chosen student has won an award for English or Math is \(\frac{56}{81}\)

Question 5