A cubical die faces marked 1, 2, 3,.... 6 tossed such that the probabi

1.	A cubical die faces marked 1, 2, 3,.... 6 tossed such that the probability of throwing the number t is proportional to t2. The probability that the number 5 has appeared given that when the die is rolled the number turned up is not even is
Answer» Given that probability of throwing a number t is proportional to t². i.e., p(t) \(\propto\) t² ⇒ p(t) = k.t², where k is proportionally constant. i.e., P(1) = k, P(2) = 4k, P(3) = 9k P(u) = 16k, P(5) = 25k & P(6) = 36 k Let event A be has appeared and event B be number turned up is not even number P(B) = P(1 or 3 or 5) = P(1) + P(3) + P(5) (\(\because\) All outcomes are independent) = k + 9k + 25k = 35k P(A \(\cap\) B) = P(5) = 25k \(\therefore\) P\((\frac{A}B)=\frac{P(A\cap B)}{P(B)}\) = \(\frac{25k}{35k}\) = \(\frac{25}{35}\) = \(\frac57\) Hence, probability that number 5 has appeared when given that the die is rolled the number turned up is not even is P(A/B) = 5/7

A cubical die faces marked 1, 2, 3,.... 6 tossed such that the probability of throwing the number t is proportional to t2. The probability that the number 5 has appeared given that when the die is rolled the number turned up is not even is

Answer»

Given that probability of throwing a number t is proportional to t².

i.e., p(t) \(\propto\) t²

⇒ p(t) = k.t², where k is proportionally constant.

i.e., P(1) = k, P(2) = 4k, P(3) = 9k

P(u) = 16k, P(5) = 25k & P(6) = 36 k

Let event A be has appeared and event B be number turned up is not even number

P(B) = P(1 or 3 or 5) = P(1) + P(3) + P(5)

(\(\because\) All outcomes are independent)

= k + 9k + 25k

= 35k

P(A \(\cap\) B) = P(5) = 25k

\(\therefore\) P\((\frac{A}B)=\frac{P(A\cap B)}{P(B)}\) = \(\frac{25k}{35k}\) = \(\frac{25}{35}\) = \(\frac57\)

Hence, probability that number 5 has appeared when given that the die is rolled the number turned up is not even is P(A/B) = 5/7

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