A' draws two cards with replacement from a pack of 52 cards and 'B' throws a pair of dice what is the chance that 'A' gets both cards of same suit and 'B' gets total of 6

Options

A. `1//144`
B. `1//4`
C. `5//144`
D. `7//144`

Answer

To solve the problem, we need to find the probability that 'A' draws two cards of the same suit and 'B' rolls a total of 6 with a pair of dice. We will calculate the probabilities for both events separately and then multiply them since they are independent events.

### Step-by-Step Solution:

**Step 1: Calculate the probability that 'A' draws two cards of the same suit.**

1. **Total number of cards in a deck**: 52 cards.
2. **Number of suits**: 4 (Hearts, Diamonds, Clubs, Spades).
3. **Number of cards in each suit**: 13 cards.

- When 'A' draws the first card, it can be of any suit. The probability of drawing any card is 1 (or 52/52).
- For the second card to be of the same suit as the first card, there are 13 cards of that suit in total.

Thus, the probability that 'A' draws two cards of the same suit is calculated as follows:

\[ P(A) = P(\text{first card}) \times P(\text{second card of the same suit}) = 1 \times \frac{13}{52} = \frac{1}{4} \]

**Step 2: Calculate the probability that 'B' rolls a total of 6 with two dice.**

1. **Total outcomes when rolling two dice**: \(6 \times 6 = 36\).
2. **Favorable outcomes for a total of 6**:
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)

This gives us a total of 5 favorable outcomes.

Thus, the probability that 'B' rolls a total of 6 is:

\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{36} \]

**Step 3: Calculate the combined probability of both events.**

Since the events are independent, we multiply their probabilities:

\[ P(\text{Both A and B}) = P(A) \times P(B) = \frac{1}{4} \times \frac{5}{36} = \frac{5}{144} \]

### Final Answer:
The probability that 'A' gets both cards of the same suit and 'B' gets a total of 6 is \(\frac{5}{144}\).

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