Multiple choice

There are ninety cards in a box numbered $10,11,12,\cdots,98, 99$. Three cards are drawn from the box one by one with replacement. The probability that product of the digits on the cards will be $12$ at least once is

  1. $1-\left(\displaystyle \frac{4}{5}\right)^{3}$
  2. $\left(\displaystyle \frac{43}{45}\right)^{3}$
  3. $1-\left(\displaystyle \frac{43}{45}\right)^{3}$
  4. $\displaystyle \frac{43}{45}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are 90 cards (10-99). The product of digits is 12 for pairs like (2,6), (6,2), (3,4), (4,3). There are 4 such numbers out of 90. Probability of getting 12 in one draw is 4/90 = 2/45. Probability of NOT getting 12 is 1 - 2/45 = 43/45. Probability of not getting 12 in 3 draws is (43/45)^3. Probability of at least one is 1 - (43/45)^3.

AI explanation

The probability of the product of digits equaling 12 in a single draw is 2/45, derived from the 4 successful cards (26, 34, 43, 62) out of 90 total cards. Therefore, the probability of failure on a single draw is 1 - 2/45 = 43/45. Because the cards are drawn with replacement, the probability of at least one success in three draws is found by subtracting the probability of three consecutive failures from one, resulting in 1 - (43/45)^3.

Keep practicing — related questions