Question

A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number less than 4 occurs on a single toss of the die, find P(E).

Answer 1

Answer:

$\frac{4}{9}$

Step-by-step explanation:

Let the probability of an even number occurring be p and that of odd be q. From the question, p = 2q.

Now, p and q are mutually exclusive events. Hence,

p + q = 1

2q + q = 1

3q = 1

q = 1/3

p = 2q = 2/3

There are 3 odd numbers and 3 even numbers on a die. The probability of each even number is p/3 = 2/9 and the probability of each odd number is q/3 = 1/9.

The event E is the event of a 3, 2 or 1 appearing. Here, we have two odd numbers and 1 even number.

$P(E) = 2\times\frac{1}{9} + 1\times\frac{2}{9} = \frac{4}{9}$

Answer 2

Answer:

P(E)=4/9

Step-by-step explanation:

We have six different numbers in a die, i.e., 1, 2, 3, 4, 5, 6. The probability of each sample point should be equal or greater than zero. Let's suppose that x is the probability that an odd number occurs, so, the probability that an even number occurs is 2x, because the die is loaded in such a way that an even number is twice as likely to occur as an odd number. Besides, the sum of the probability associated to an odd number with the probability associated to an even number should be equal to 1/3, because the sum of the probabilities of all sample points should be equal to 1 and we have three odd numbers and three even numbers in a die. Therefore x+2x=1/3, i.e., 3x=1/3, x = 1/9, i.e., the probability that an odd number occurs is 1/9 and the probability that an even number occurs is 2/9. E is the event that a number less than 4 occurs on a single toss of the die, so, P(E) = P(1) + P(2) + P(3) = 1/9 + 2/9 + 1/9 = 4/9.