Question

There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X > 3) is:

• A. \(\frac{14}{15}\)
• B. \(\frac{1}{15}\)
• C. \(\frac{11}{12}\)
• D. \(\frac{1}{12}\)

Hint:

In problems involving combinations and probability, it's important to first calculate the total number of possible outcomes and then determine the number of favorable outcomes. To find the probability, use the ratio of favorable outcomes to total outcomes. In this case, identifying the unfavorable outcomes (those where the sum is less than or equal to 3) simplifies the problem and leads to a straightforward calculation of the probability.

Answer 1

The Correct Option is A

There are \(\binom{6}{2} = 15\) total ways to select 2 cards from 6. The event \(X > 3\) signifies that the sum of the numbers on the two selected cards exceeds 3. The sole combination resulting in a sum less than or equal to 3 is the pair \((1, 2)\), which can be chosen in 1 way.

Therefore, the count of successful outcomes for the event \(X > 3\) is calculated as:
\(15 - 1 = 14.\)

The probability of this event is:
\(P(X > 3) = \frac{14}{15}.\)