Question

Two cards are drawn from a full pack of 52 cards at random. The probability of getting a heart and a diamond is (rounded off to two decimal places).

Hint:

For drawing two cards without replacement, calculate the probability of each event and then add the probabilities of the two possible outcomes.

Answer 1

A standard deck of cards contains 52 cards, with 13 cards in each of the 4 suits: hearts, diamonds, clubs, and spades.
We are asked to find the probability of drawing one heart and one diamond.
Step 1: Find the probability of drawing a heart first and a diamond second.
- The probability of drawing a heart first is: \[ P(\text{Heart first}) = \frac{13}{52} = \frac{1}{4} \] - After drawing the heart, there are now 51 cards left, with 13 diamonds remaining. The probability of drawing a diamond second is: \[ P(\text{Diamond second}) = \frac{13}{51} \] The combined probability of drawing a heart first and a diamond second is: \[ P(\text{Heart first and Diamond second}) = \frac{1}{4} \times \frac{13}{51} = \frac{13}{204} \] Step 2: Consider the reverse case, where a diamond is drawn first and a heart second.
- The probability of drawing a diamond first is: \[ P(\text{Diamond first}) = \frac{13}{52} = \frac{1}{4} \] - After drawing the diamond, there are now 51 cards left, with 13 hearts remaining. The probability of drawing a heart second is: \[ P(\text{Heart second}) = \frac{13}{51} \] The combined probability of drawing a diamond first and a heart second is also: \[ P(\text{Diamond first and Heart second}) = \frac{1}{4} \times \frac{13}{51} = \frac{13}{204} \] Step 3: Add the probabilities of both cases to get the total probability: \[ P(\text{Heart and Diamond}) = \frac{13}{204} + \frac{13}{204} = \frac{26}{204} = \frac{13}{102} \approx 0.1275 \] Final Answer: \[ \boxed{0.13} \]