Question

1. a teacher has 8 cards in a bag with different letters and numbers as shown. she asks two students to pick a card out of the bag without replacing them.

a 5 1 7
1 2 3 k

a. what is the probability that both students pick 1?
b. what is the probability that neither student picks a letter?
c. what is the probability that the first student picks k and the second student picks an odd number?
d. what is the probability that the first student picks a letter and the second student picks a prime number?

Explanation:

Step1: Calculate probability for part a

The probability that the first student picks 1 is $\frac{2}{8}$ since there are 2 cards with 1 out of 8 cards. After the first - student picks a 1, there is 1 card with 1 left out of 7 cards. So the probability that the second student picks 1 is $\frac{1}{7}$. The probability that both students pick 1 is $\frac{2}{8}\times\frac{1}{7}=\frac{2}{56}=\frac{1}{28}$.

Step2: Calculate probability for part b

There are 6 non - letter cards out of 8. The probability that the first student picks a non - letter card is $\frac{6}{8}$. After the first student picks a non - letter card, there are 5 non - letter cards left out of 7. So the probability that the second student also picks a non - letter card is $\frac{5}{7}$. The probability that neither student picks a letter is $\frac{6}{8}\times\frac{5}{7}=\frac{30}{56}=\frac{15}{28}$.

Step3: Calculate probability for part c

The probability that the first student picks K is $\frac{1}{8}$. After the first student picks K, there are 5 odd - numbered cards out of 7 remaining cards. So the probability that the second student picks an odd number is $\frac{5}{7}$. The probability that the first student picks K and the second student picks an odd number is $\frac{1}{8}\times\frac{5}{7}=\frac{5}{56}$.

Step4: Calculate probability for part d

The probability that the first student picks a letter is $\frac{2}{8}$. After the first student picks a letter, there are 4 prime - numbered cards (2, 3, 5, 7) out of 7 remaining cards. So the probability that the second student picks a prime number is $\frac{4}{7}$. The probability that the first student picks a letter and the second student picks a prime number is $\frac{2}{8}\times\frac{4}{7}=\frac{8}{56}=\frac{1}{7}$.

Answer:

a. $\frac{1}{28}$
b. $\frac{15}{28}$
c. $\frac{5}{56}$
d. $\frac{1}{7}$