Answered

  1. A bag contains four 2007 pennies, two 2005 pennies, and six 2001 pennies.
    A student removes 1 penny without looking, records the year, then
    returns
    the penny to the bag. The process is repeated.
    Find the probability of each event:
    a) two 2001 pennies
    b) a 2007 penny, then a 2005 penny
    c) a 2001 penny, then a 2005 penny
    d) not a 2001 penny, then a 2007 penny

Verified Answer :

GinnyAnswer

Sure, let's carefully work through each part of the problem to find the required probabilities.

### Preliminary Calculation
First, we need the total number of pennies in the bag:
- Four 2007 pennies
- Two 2005 pennies
- Six 2001 pennies

The total number of pennies is:
\(\[ 4 + 2 + 6 = 12 \]\)

### a) Probability of drawing two 2001 pennies

To find the probability of drawing two 2001 pennies in a row (with replacement), we calculate the probability of drawing a 2001 penny on the first draw, and then again on the second draw:
- Probability of drawing a 2001 penny first:
\(\[ \frac{6}{12} = 0.5 \]\)
- Probability of drawing another 2001 penny:
\(\[ \frac{6}{12} = 0.5 \]\)

Since these events are independent (due to replacement), their combined probability is:
\(\[ 0.5 \times 0.5 = 0.25 \]\)

So, the probability is:
\(\[ \boxed{0.25} \]\)

### b) Probability of drawing a 2007 penny, then a 2005 penny

To find the probability of drawing a 2007 penny first and a 2005 penny second:
- Probability of drawing a 2007 penny first:
\(\[ \frac{4}{12} = \frac{1}{3} \]\)
- Probability of drawing a 2005 penny second:
\(\[ \frac{2}{12} = \frac{1}{6} \]\)

Their combined probability is:
\(\[ \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \approx 0.0556 \]\)

So, the probability is:
\(\[ \boxed{0.0556} \]\)

### c) Probability of drawing a 2001 penny, then a 2005 penny

To find the probability of drawing a 2001 penny first and a 2005 penny second:
- Probability of drawing a 2001 penny first:
\(\[ \frac{6}{12} = 0.5 \]\)
- Probability of drawing a 2005 penny second:
\(\[ \frac{2}{12} = \frac{1}{6} \]\)

Their combined probability is:
\(\[ 0.5 \times \frac{1}{6} = \frac{1}{12} \approx 0.0833 \]\)

So, the probability is:
\(\[ \boxed{0.0833} \]\)

### d) Probability of not drawing a 2001 penny, then drawing a 2007 penny

To find the probability of not drawing a 2001 penny first and then drawing a 2007 penny:
- Probability of not drawing a 2001 penny first (i.e., drawing a 2007 or 2005 penny):
\(\[ \frac{4 (2007) + 2 (2005)}{12} = \frac{6}{12} = 0.5 \]\)
- Probability of drawing a 2007 penny second:
\(\[ \frac{4}{12} = \frac{1}{3} \]\)

Their combined probability is:
\(\[ 0.5 \times \frac{1}{3} = \frac{1}{6} \approx 0.1667 \]\)

So, the probability is:
\(\[ \boxed{0.1667} \]\)

This step-by-step solution provides the probabilities for each event in the given problem.

toowaiting71

Answer:

To find the probabilities, we first need to determine the total number of pennies in the bag, and then calculate the probabilities based on the given information.

Total number of pennies in the bag: \(4 + 2 + 6 = 12\)

a) Probability of drawing two 2001 pennies:

There are 6 2001 pennies in the bag, so the probability of drawing one 2001 penny on the first draw is \( \frac{6}{12} = \frac{1}{2} \). After replacing the first penny, the probability of drawing another 2001 penny is still \( \frac{6}{12} = \frac{1}{2} \). So, the probability of drawing two 2001 pennies in a row is:

\[ P(\text{two 2001 pennies}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

b) Probability of drawing a 2007 penny, then a 2005 penny:

For the first draw, there are 4 2007 pennies in the bag, so the probability of drawing one 2007 penny is \( \frac{4}{12} = \frac{1}{3} \). After replacing the first penny, there are 2 2005 pennies in the bag, so the probability of drawing one 2005 penny is \( \frac{2}{12} = \frac{1}{6} \). So, the probability of drawing a 2007 penny followed by a 2005 penny is:

\[ P(\text{2007, then 2005}) = \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \]

c) Probability of drawing a 2001 penny, then a 2005 penny:

For the first draw, there are 6 2001 pennies in the bag, so the probability of drawing one 2001 penny is \( \frac{6}{12} = \frac{1}{2} \). After replacing the first penny, there are still 2 2005 pennies in the bag, so the probability of drawing one 2005 penny is \( \frac{2}{12} = \frac{1}{6} \). So, the probability of drawing a 2001 penny followed by a 2005 penny is:

\[ P(\text{2001, then 2005}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \]

d) Probability of not drawing a 2001 penny, then drawing a 2007 penny:

The probability of not drawing a 2001 penny on the first draw is \(1 - \frac{6}{12} = \frac{1}{2} \), and the probability of drawing a 2007 penny on the second draw is \( \frac{4}{12} = \frac{1}{3} \). So, the probability of not drawing a 2001 penny followed by drawing a 2007 penny is:

\[ P(\text{not 2001, then 2007}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \]

Therefore:

a) \( P(\text{two 2001 pennies}) = \frac{1}{4} \)

b) \( P(\text{2007, then 2005}) = \frac{1}{18} \)

c) \( P(\text{2001, then 2005}) = \frac{1}{12} \)

d) \( P(\text{not 2001, then 2007}) = \frac{1}{6} \)

Step-by-step explanation:

Related Questions