Question

A bag of coins contains 6 quarters, 4 dimes, 5 nickels, and 7 pennies. Determine the probability of drawing a dime then a quarter, without replacement.
A. 12/231

B. 5/11

C. 6/121

D. 4/77

(There will most likely be more so if you want easy points just check)

Answer 1

Answer:

i think the answer is B

Step-by-step explanation:

Answer 2

Answer:  The correct option is

(D) $\dfrac{4}{77}.$

Step-by-step explanation:  Given that a bag of coins contains 6 quarters, 4 dimes, 5 nickels, and 7 pennies.

We are given to find the probability of drawing a dime then a quarter, without replacement.

Total number of coins = 6+4+5+7 = 22.

Let A represents the event of drawing a dime and B represents the event of drawing a quarter without replacement after drawing a dime.

Then, the probabilities of A and B are

$P(A)=\dfrac{\textup{number of dimes}}{\textup{total number of coins}}=\dfrac{4}{22}=\dfrac{2}{11}$

and

$P(B)=\dfrac{\textup{total number of quarters}}{\textup{total number of coins left after drawing a dime}}=\dfrac{6}{21}=\dfrac{2}{7}.$

Therefore, the probability of drawing a dime then a quarter, without replacement is given by

$P(A\cap B)\\\\=P(A)\times P(B)~~~~~~~~~~~~~~~~~~~[\textup{since A and B are independent events}]\\\\\\=\dfrac{2}{11}\times\dfrac{2}{7}\\\\\\=\dfrac{4}{77}.$

Thus, the required probability is $\dfrac{4}{77}.$

Option (D) is CORRECT.