Question

If a game consists of picking a number from a bag of beans numbered 2, 2, 3, 4, and 5 and flipping a coin, which probabilities are equivalent?


I.
II.
III.
A. I and II
B. I and III
C. II and III
D. none of the above

Answer 1

The true statement is (d) none of the above

Missing information

I. P(3 and tail)

II. P(even and head)

III .P(odd and head)​

How to determine the equivalent probabilities?

We start by calculating each probability

I. P(3 and tail)

There is only one 3 in the 5 numbers.

So:

P(3) = 1/5

The probability of a tail in a coin is:

P(Coin) = 1/2

So, we have:

P(3 and tail) = 1/5 * 1/2

P(3 and tail) = 1/10

II. P(even and head)

There are three even numbers in the 5 numbers.

So:

P(even) = 3/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(even and head) = 3/5 * 1/2

P(even and head) = 3/10

III .P(odd and head)​

There are two odd numbers in the 5 numbers.

So:

P(odd) = 2/5

The probability of a head in a coin is:

P(head) = 1/2

So, we have:

P(odd and head) = 2/5 * 1/2

P(even and head) = 1/5

By comparing the results, none of the probabilities are equal.

Hence, the true statement is (d) none of the above

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