A nickel, a penny, and a dime are placed inside a bag. One coin is randomly picked three times. If the coin is replaced after each draw, what is the probability that a dime is picked exactly twice?(1 po int)

Responses

19
Start Fraction 1 over 9 End Fraction

29
Start Fraction 2 over 9 End Fraction

23
Start Fraction 2 over 3 End Fraction

727

1 answer

To find the probability of picking a dime exactly twice when randomly drawing from the bag three times (with replacement), we can use the binomial probability formula:

\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]

Where:

  • \( n = 3 \) (the number of trials)
  • \( k = 2 \) (the number of successes, which in this case is picking a dime)
  • \( p \) is the probability of success on an individual trial.

First, we identify the probabilities:

  • The total number of coins = 3 (a nickel, a penny, and a dime).
  • The probability of picking a dime \( p = \frac{1}{3} \).
  • The probability of not picking a dime \( 1-p = \frac{2}{3} \).

Now we substitute the values into the binomial formula:

\[ P(X=2) = \binom{3}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{3-2} \]

Calculating the binomial coefficient:

\[ \binom{3}{2} = 3 \]

Now substituting all the values:

\[ P(X=2) = 3 \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{1} \]

Calculating each term:

\[ = 3 \cdot \frac{1}{9} \cdot \frac{2}{3} \] \[ = 3 \cdot \frac{2}{27} \] \[ = \frac{6}{27} \] \[ = \frac{2}{9} \]

Thus, the probability that a dime is picked exactly twice is:

\[ \frac{2}{9} \]

The correct answer is:

Start Fraction 2 over 9 End Fraction.