Chapter 5: Q T5.11. (page 337)

Your teacher has invented a “fair” dice game to play. Here’s how it works. Your teacher will roll one fair eight-sided die, and you will roll a fair six-sided die. Each player rolls once, and the winner is the person with the higher number. In case of a tie, neither player wins. The table shows the sample space of this chance process.
(a) Let A be the event “your teacher wins.” Find P(A).
(b) Let B be the event “you get a $3$ on your first roll.” Find P(A ∪ B).
(c) Are events A and B independent? Justify your answer.

Short Answer

Expert verified

Part (a) The probability is $0.5625$

Part (b) The probability is $0.625$

Part (c) No.

Step by step solution

Part (a) Step 1. Given

The table is

Part (a) Step 2. Concept used

Probability= $\frac{F a v o u r a b l e o u t c o m e s}{T o t a l o u t c o m e s}$

Part (a) Step 3. Calculation

The teacher is victorious in event A.

In $1 + 2 + 3 + 4 + 5 + 6 + 6 = 27$ situations, the teacher will win.

There are $48$ examples in total.

P (A) is computed as:

P (A) = \frac{27}{48} = 0.5625

Thus, $P (A) = 0.5625$

Part (b) Step 1. Calculation

The event B indicates that on the first roll, a $3$ will be achieved.

$P (A \cup B)$ is computed as:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

$P (A \cup B) = \frac{27}{48} + \frac{8}{48} - \frac{5}{48} = \frac{30}{48} = 0.625$

Thus, the probability is $0.625$

Part (c) Step 1. Calculation

$P (A) = \frac{27}{48} P (B) = \frac{8}{48} P (C) = \frac{5}{48}$

The following conditions must be met for the occurrences to be independent:

$P (A \cap B) = P (A) \times P (B) \frac{27}{48} \times \frac{7}{48} = \frac{5}{48} \frac{216}{2304} i s n o t e q u a l t o \frac{5}{48}$

Because the preceding criteria are not met, A and B are not independent events.

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Short Answer

Step by step solution

Part (a) Step 1. Given

Part (a) Step 2. Concept used

Part (a) Step 3. Calculation

Part (b) Step 1. Calculation

Part (c) Step 1. Calculation

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