Question

a bag contains 10 equal - sized pieces of paper, numbered 1 through 10. you draw one piece of paper, do not put it back in the bag, and then draw another slip of paper. what is the probability that you will pick out the number 4 on your first draw and the number 3 on your second draw?
options:
a. \\(\frac{1}{10}\\)
b. \\(\frac{1}{100}\\)
c. \\(\frac{1}{19}\\)
d. \\(\frac{1}{90}\\)

Explanation:

Step1: Probability of first draw

There are 10 papers, so the probability of drawing 4 first is $\frac{1}{10}$.

Step2: Probability of second draw

After drawing 4, 9 papers remain. Probability of drawing 3 next is $\frac{1}{9}$.

Step3: Multiply probabilities

Multiply the two probabilities: $\frac{1}{10} \times \frac{1}{9} = \frac{1}{90}$. Wait, no, wait—wait, the options have $\frac{1}{90}$? Wait, let's check again. Wait, the first draw: 10 papers, chance of 4 is 1/10. Second draw: 9 papers left, chance of 3 is 1/9. So 1/10 * 1/9 = 1/90. But let's check the options. Wait, the last option (D?) is 1/90? Wait, the user's image: the last option is 1/90? So the correct answer is the option with 1/90.

Answer:

The option with $\boldsymbol{\frac{1}{90}}$ (e.g., if it's option D, then D. $\frac{1}{90}$)