help

[wind bane]

25 cards are marked from 1 to 25 and 2 are drawn at random. find the probability that one card drawn is multiple of 4 and another one is a multiple of 6

[thin kite]

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[quiet void]

Let's break this down step by step:

1. Total number of cards: 25
2. Total number of ways to draw 2 cards: 25C2 = 25!/(2! × 23!) = 300
3. Number of cards that are multiples of 4: 6 (4, 8, 12, 16, 20, 24)
4. Number of cards that are multiples of 6: 4 (6, 12, 18, 24)
5. Note that two cards (12 and 24) are common to both sets.

Now, let's find the probability:

a. Probability of drawing a multiple of 4: 6/25
b. Probability of drawing a multiple of 6: 4/25
c. Probability of drawing both (assuming independence): (6/25) × (4/25) = 24/625

However, we need to consider the overlap (cards 12 and 24). We've counted these twice, so we need to subtract the probability of drawing both from the multiples of 4 and 6:

d. Probability of drawing both (corrected): 24/625 - 2/625 = 22/625

So, the probability that one card drawn is a multiple of 4 and another is a multiple of 6 is 22/625.

[wind bane]

thank you

[jagged jacinth]

The denominator should not be 25^2, since the cards are drawn without replacement

You should keep using 300 from your second point as the denominator, and consider the number of combinations which fit the problem setting

[jagged jacinth]

consider the number of combinations which fit the problem setting
One way of counting such combinations is to split the cards into disjoint sets
Number of cards that are multiples of 4 and not multiples of 6 = 4
Number of cards that are multiples of 6 and not multiples of 4 = 2
Number of cards that are multiples of both 4 and 6 = 2
Number of desired combinations = 4 x 2 + 4 x 2 + 2 x 2 + 2C2 = 21

[quiet void]

You're absolutely right! I apologize for my earlier mistake. Your approach is a great way to count the combinations that fit the problem setting.

Let's break it down:

• Number of cards that are multiples of 4 and not multiples of 6: 4 (8, 16, 20, 24)
• Number of cards that are multiples of 6 and not multiples of 4: 2 (6, 18)
• Number of cards that are multiples of both 4 and 6: 2 (12, 24)
• Number of desired combinations:
  • Multiples of 4 and not 6 with multiples of 6 and not 4: 4 × 2 = 8
  • Multiples of 4 and not 6 with multiples of both 4 and 6: 4 × 2 = 8
  • Multiples of 6 and not 4 with multiples of both 4 and 6: 2 × 2 = 4
  • Both multiples of both 4 and 6: 2C2 = 1
    Total: 8 + 8 + 4 + 1 = 21

So, there are 21 combinations that fit the problem setting.

Now, let's calculate the probability:

Probability = Number of desired combinations / Total number of combinations
= 21 / 300
= 7 / 100

So, the probability that one card drawn is a multiple of 4 and another is a multiple of 6 is 7/100. Thank you for your response