Question

1. mazen has exams coming up in both english and chemistry. he gives himself 11:7 odds in favour of passing the english test and 2:3 odds against passing the chemistry test. if these are independent events, what are the odds in favour of mazen passing both of these tests?
2. calculate the probabilities of each event happening:

a) rolling a sum less than 6 with two die.
b) rolling a sum less than 6 with two die,
if you knew the first die rolled was a 4
c) explain how including the condition changes the probability of rolling a sum less than 6.

Explanation:

Step1: Calculate probability of passing English

Odds in favour of passing English is \(11:7\). Probability of passing English \(P(E)=\frac{11}{11 + 7}=\frac{11}{18}\)

Step2: Calculate probability of passing Chemistry

Odds against passing Chemistry is \(2:3\). Odds in favour of passing Chemistry is \(3:2\). Probability of passing Chemistry \(P(C)=\frac{3}{3+2}=\frac{3}{5}\)

Step3: Calculate probability of passing both

Since events are independent, \(P(E\cap C)=P(E)\times P(C)=\frac{11}{18}\times\frac{3}{5}=\frac{11}{30}\)
Probability of not passing both \(P(\overline{E\cap C})=1 - \frac{11}{30}=\frac{19}{30}\)

Step4: Calculate odds in favour

Odds in favour of passing both \(=\frac{P(E\cap C)}{P(\overline{E\cap C})}=\frac{\frac{11}{30}}{\frac{19}{30}}=\frac{11}{19}\) or \(11:19\)

Answer:

The odds in favour of Mazen passing both tests is \(11:19\)