Question

A biased die is rolled such that the probability of getting k dots,1≤k≤6, on the upper face of the die is proportional to k. Then the probability that five dots appear on the upper face of the die is

• A. $\dfrac{16}{21}$
• B. $\dfrac{1}{21}$
• C. $\dfrac{2}{21}$
• D. $\dfrac{3}{21}$
• E. $\dfrac{5}{21}$

Answer 1

Answer: (E)

$\dfrac{5}{21}$

Answer 2

Given data:

The probability of getting $k$ dots on the die, where $1 ≤ k ≤ 6.$

Since, upper face of the die is proportional to $k,$

Then, $P(k)∝k$

$⇒P(1) = k × C$, (where C is the constant of proportionality.)

$P(2) = 2k × C$

$P(3) = 3k × C$

$P(4) = 4k × C$

$P(5) = 5k × C$

$P(6) = 6k × C$

Now, to find the value of $C$, (we know fact that the sum of all probabilities must be 1)

Hence,

$P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$

$⇒(k × C) + (2k × C) + (3k × C) + (4k × C) + (5k × C) + (6k × C) = 1$

$⇒21k × C = 1$

$⇒C = \dfrac{1}{(21k)}$

Now probability of getting 5 dots will be,

$P(5) = 5k × C = 5k × \dfrac{1 }{21k} = \dfrac{5}{21}$ (_Ans.)