Question

9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions

Answer 1

Answer:

$\frac{7}{30}\approx0.23$

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

$P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5$

Now, probability of getting the second question as easy question is given as:

$P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}$

Now, probability that the first two contestants will get easy questions is given by the product of $P(E1)\ and\ P(E2)$. So,

$P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23$

Therefore, the probability that the first two contestants will get easy questions is $\frac{7}{30}\approx0.23$