Question

Two dice D1 and D2 has 4 red and 2 blue and 3 red and 3 blue faces respectively. A card is drawn from an ordinary pack of 52 playing cards. If it shows heart or king, te game continues by throwing D1 and if drawn crd is a face card D2 is thrown. prob of getting red face on first tryis ? and if its given that red face appeared on first throw, then probability that drawn card is an ace

Answer 1

1. Probability of red face on first try: $\frac{41}{156}$.
2. Given a red face appeared, probability that the drawn card is an ace: $\frac{2}{41}$.

Answer 2

Solution:

1. Step 1. Partition the card outcomes:

• Cards giving D₁ throw:
  Condition: Card is a heart or a king.

  • Hearts: 13 cards
  • Kings (non‐hearts): 3 cards
    → Total = 13 + 3 = 16 cards.
    So,
  $$P(D_1\ \text{event}) = \frac{16}{52}.$$

• Cards giving D₂ throw:
  Condition: Card is a face card (J, Q, K) but not already counted in D₁.
  Total face cards in a deck = 12. In hearts, face cards (J, Q, K of hearts) = 3 (already in D₁) and for the other suits, note that the king in each suit is already in D₁.
  For non‐heart face cards, each suit has 3 face cards; for spades, clubs, diamonds that gives 3×3 = 9, but each suit’s king (3 in total) must be removed.
  So remaining = 9 – 3 = 6 cards.
  Thus,

  $$P(D_2\ \text{event}) = \frac{6}{52}.$$

(For other cards, no dice is thrown.)

2. Step 2. Dice outcomes:

• Die D₁: Faces: 4 red, 2 blue → $$P(\text{red on }D_1) = \frac{4}{6} = \frac{2}{3}.$$
• Die D₂: Faces: 3 red, 3 blue → $$P(\text{red on }D_2) = \frac{3}{6} = \frac{1}{2}.$$

3. Step 3. Overall probability for a red face on the first try:

Only when a dice is thrown (via D₁ or D₂) can a red face appear:

$$P(\text{red face on first try}) = P(D_1)\cdot P(\text{red on }D_1) + P(D_2)\cdot P(\text{red on }D_2)$$

Substitute values:

$$= \frac{16}{52}\cdot\frac{2}{3} + \frac{6}{52}\cdot\frac{1}{2} = \frac{32}{156} + \frac{9}{156} = \frac{41}{156}.$$

4. Step 4. Conditional probability that the drawn card is an ace given a red face:

Only an ace can come from the hearts group (since aces from other suits do not qualify for either event).

• In hearts, there is 1 ace (Ace of hearts).
• When Ace of hearts is drawn, it falls into the D₁ event. The probability that the die shows red in that case is $\frac{2}{3}$.
  Thus,

$$P(\text{Ace of hearts and red}) = \frac{1}{52} \cdot \frac{2}{3} = \frac{2}{156}.$$

Now, using conditional probability:

$$P(\text{card is Ace} \mid \text{red face}) = \frac{P(\text{Ace and red})}{P(\text{red face})} = \frac{\frac{2}{156}}{\frac{41}{156}} = \frac{2}{41}.$$

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Mermaid Diagram of the Process:

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Final Answers:

1. Probability of red face on first try: $\frac{41}{156}$.
2. Given a red face appeared, probability that the drawn card is an ace: $\frac{2}{41}$.

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Summary:

• Explanation (minimal):
  Count cards that trigger D₁ (16/52) and D₂ (6/52). Multiply by corresponding dice red face probabilities (D₁: $\frac{2}{3}$, D₂: $\frac{1}{2}$) and add. For the conditional probability, only Ace of hearts leads to a dice throw (D₁) so compute its combined probability and condition on the total red outcome.

• Answer:
  (1) $\frac{41}{156}$
  (2) $\frac{2}{41}$