Q5 of 43 Page 296

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Given that probability of odd numbers

= 2 × (Probability of even number)


⇒ P (Odd) = 2 × P (Even)


Now, P (Odd) + P (Even) = 1


⇒ 2P(Even) + P(Even) = 1


⇒ 3P(Even) = 1



So,


Now, Total number occurs on a single roll of die = 6


and the number greater than 3 = 4, 5 or 6


So, P(G) = P(number greater than 3)


= P(number is 4, 5 or 6)


Here, 4 and 6 are even numbers and 5 is odd


∴ P(G) = 2 × P(Even) × P(Odd)




Hence, the required probability is


More from this chapter

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Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

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An experiment consists of rolling a die until a 2 appears.

(i) How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die?


(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die?

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In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

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If A and B are mutually exclusive events, P (A) = 0.35 and P (B) = 0.45, find

(a) P (A′)


(b) P (B′)


(c) P (A ∪ B)


(d) P (A ∩ B)


(e) P (A ∩ B′)


(f) P (A′∩ B′)