Forty cards are placed into a box, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four cards. Four cards are drawn from the box at random and without replacement. Let $p$ be the probability that all four cards bear the same number. Let $q$ be the probability that three of the cards bear a number $a$ and the other bears a number $b$ that is not equal to $a$. What is the value of $q/p$?

The value of q/p = 144

Step-by-step explanation:

Number of cards in the box = 40

Each bearing a number: 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10

Each number appears 4 times cards in total

p = the probability that all four cards bear the same number.

q = the probability that three of the cards bear a number 'a' and the other bears a number 'b' that is not equal to 'a'.

The cards were chosen without replacement.

For Probability without replacement, the total number of items decrease after each pick. When considering same items, the number of the same item also decrease after each pick.

a) In this question, the order of the 4 cards picked is irrelevant.

Pr (4 same cards) = p = 4/40 × 3/39 × 2/38 × 1/37

p = 24/2193360 = 1/91390

Pr (3same cards and 1 different card) = q

Probability of picking 'a' card = 4/40

Probability of picking other cards aside 'a' = Probability of not picking 'a' card

= 36/40

Since we were not told what the particular number of the other number is, it would be any of the remaining 36 numbers.

b) In this question, we would consider the order of the 4 cards picked.

q = Pr(baaa) + Pr(abaa) + Pr(aaba) + Pr(aaab)

Without replacement

q = (36/40 × 4/39 × 3/38 × 2/37) + (4/40 × 36/39 × 3/38 × 2/37) + (4/40 × 3/39 × 36/38 × 2/37) + (4/40 × 3/39 × 2/38 × 36/37)

q = 4[(36×24)/2193360]

q= 144(24)/2193360 = 144(24/2193360)

q= 144(1/91390) = 144/91390

The value of q/p = (144/91390)/(1/91390)

The value of q/p = (144/91390) × (91390/1)

The value of q/p = 144

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