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Mekhanik [1.2K]
2 years ago
6

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

ree of its faces, and the number 3 painted on one of its faces. Assume that each face is equally likely to come up. If the die were loaded so that the face with the 3 on it were twice as likely to come up as each of the other five faces, would this change the value of P(odd number)
Mathematics
1 answer:
grigory [225]2 years ago
7 0

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

S = \{1,1,2,2,2,3\}

And the probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

P(1) = \frac{1}{3}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

Take LCM

P(Odd\ Number) = \frac{2+1}{6}

P(Odd\ Number) = \frac{3}{6}

P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

P(Odd\ Number) = P(1) + P(3)

P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}

Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

P(Odd\ Number) =  \frac{1}{2} --- Before

P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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