Probability of Card 1 Selection

The problem asks for the probability that card number 1 is among the four cards drawn from a pack containing 6 cards numbered 1, 2, 3, 4, 5, and 6. The drawing is done one by one without replacement.

We are drawing 4 cards out of a total of 6 cards. We want to find the probability that the specific card numbered 1 is included in the set of 4 cards that are drawn.

Consider the 6 cards as distinct items. When we draw 4 cards without replacement, we are essentially selecting a subset of 4 cards from the 6. Each card in the original pack of 6 has an equal chance of being part of the selected group of 4 cards.

Out of the 6 cards available, 4 cards are selected. The probability that any particular card (like card number 1) is among the selected cards is the ratio of the number of cards selected to the total number of cards available.

The total number of cards in the pack is 6.

The number of cards drawn is 4.

The probability that card number 1 is drawn in the selected four cards can be directly calculated using the ratio:

Probability (Card 1 is drawn) = \(\frac{\text{Number of cards drawn}}{\text{Total number of cards}}\)

Probability (Card 1 is drawn) = \(\frac{4}{6}\)

This fraction can be simplified to \(\frac{2}{3}\), but the option is given as \(\frac{4}{6}\).

Let's consider an alternative way to think about this using combinations, although the direct ratio method is simpler here:

• Total number of ways to choose 4 cards from 6 is given by the combination formula \({}^nC_r = \frac{n!}{r!(n-r)!}\).
  Total combinations = \({}^6C_4 = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15\).
• Number of ways to choose 4 cards such that card 1 is included: We must select card 1 (\({}^1C_1 = 1\)), and then select the remaining 3 cards from the other 5 cards (\({}^5C_3 = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10\)).
  Favorable combinations = \({}^1C_1 \times {}^5C_3 = 1 \times 10 = 10\).
• The probability is the ratio of favorable combinations to total combinations.
  Probability = \(\frac{10}{15} = \frac{2}{3}\).

Both methods yield the probability \(\frac{2}{3}\), which is equivalent to \(\frac{4}{6}\).

Thus, the probability that the card numbered 1 is drawn in the selected four cards is \(\frac{4}{6}\).