ten cards are numbered \1\ through \10\ and placed in a bag. one card is chosen and replaced after noting…

ten cards are numbered \1\ through \10\ and placed in a bag. one card is chosen and replaced after noting its number. a second card is then drawn. what is the probability that the first card is \1\ and the second card is \10\?\na $\frac{1}{50}$\nb $\frac{1}{5}$\nc $\frac{1}{100}$\nd $\frac{1}{90}$

ten cards are numbered \1\ through \10\ and placed in a bag. one card is chosen and replaced after noting its number. a second card is then drawn. what is the probability that the first card is \1\ and the second card is \10\?\na $\frac{1}{50}$\nb $\frac{1}{5}$\nc $\frac{1}{100}$\nd $\frac{1}{90}$

Answer

Explanation:

Step1: Calculate probability of first - card draw

The probability of drawing the card numbered "1" on the first draw. There are 10 cards in total, so the probability $P(1)$ of drawing the "1" card is $\frac{1}{10}$ since there is 1 card numbered "1" out of 10 cards.

Step2: Calculate probability of second - card draw

The probability of drawing the card numbered "10" on the second draw. Since the first card is replaced, there are still 10 cards in total. So the probability $P(10)$ of drawing the "10" card is $\frac{1}{10}$ as there is 1 card numbered "10" out of 10 cards.

Step3: Calculate joint probability

Since the two draws are independent events (because of replacement), the probability that the first card is "1" and the second card is "10" is the product of the probabilities of each event. So $P = P(1)\times P(10)=\frac{1}{10}\times\frac{1}{10}=\frac{1}{100}$.

Answer:

C. $\frac{1}{100}$