Consider a e with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is .
0.142

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Solution

The correct option is A 0.142
Given:
$P(n dots) \propto n$
$P(n dots) = k n$
By total probability, we have

$\sum p_{i} = 1$
$\Rightarrow k + 2 k + 3 k + 4 k + 5 k + 6 k = 1$

$\Rightarrow k = \frac{1}{21}$

$∴$ Probability of the face with three dots showing up
$= 3 k = \frac{3}{21} = 0.142.$

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Consider a e with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is .0.142

The correct option is A 0.142Given: P(n dots)∝n P(n dots)=kn By total probability, we have ∑pi=1 ⇒ k+2k+3k+4k+5k+6k=1 ⇒ k=121 ∴ Probability of the face with three dots showing up =3k=321=0.142.

Consider a e with the property that the probability of a face with n dots showing up is proportional to n. The probability of the face with three dots showing up is .
0.142

The correct option is A 0.142
Given:
$P(n dots) \propto n$
$P(n dots) = k n$
By total probability, we have

$\sum p_{i} = 1$
$\Rightarrow k + 2 k + 3 k + 4 k + 5 k + 6 k = 1$

$\Rightarrow k = \frac{1}{21}$

$∴$ Probability of the face with three dots showing up
$= 3 k = \frac{3}{21} = 0.142.$