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zaharov [31]
2 years ago
5

There are 4 quarters, 5 nickels and 3 dimes in a jar. One coin is randomly drawn,

Mathematics
1 answer:
vagabundo [1.1K]2 years ago
3 0

Answer:

\displaystyle\frac{5}{36}

Step-by-step explanation:

Probability is used to find how likely an outcome is to happen. Probability can be expressed as a fraction with the total outcomes as the denominator and the number of successful outcomes (what you want to happen) as the numerator.

Sample Size

The sample size is the number of possible outcomes. In this case, the sample size will be the total number of coins. To find the sample size we need to add together all of the coins.

  • 4 + 5 + 3 = 12

This means that the sample size is 12. For this type of probability, the sample size will serve as the denominator for each probability.

Replacement

In the question, it is stated that coins are replaced. This means that the sample size will not change at any point. So, the denominator will be the same for every probability.

Creating Probability Fractions

We are looking for the probability of getting a quarter and nickel, so we need to find the fractions that represent both situations.

First, let's find the quarter. As stated in the first paragraph, the number of successful outcomes is the numerator. There are 4 quarters, so 4 is the number of successful outcomes. Therefore, 4 will be the numerator. Then, 12 will be the denominator because 12 is the sample size.

  • \displaystyle\frac{4}{12}

Next, we need to find the probability of a nickel. Since there are 5 nickels, there are 5 successful outcomes. Then, the sample size is still 12 because there is replacement.

  • \displaystyle\frac{5}{12}

Complex Probability

Now we have the probability of both a quarter and nickel alone. However, this question asks about the probability of both, so this is an example of complex probability. To find complex probability, you have to multiply each probability together.

  • \displaystyle\frac{4}{12} *\frac{5}{12}=\frac{20}{144}

To reach the final answer we have to simplify the answer.

  • \displaystyle\frac{20}{144} =\frac{5}{36}

This means that the probability of pulling a quarter and then a nickel is 5/36.

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Answer:

Part A:

The probability that all of the balls selected are white:

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

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Step-by-step explanation:

A is the event all balls are white.

D_i is the dice outcome.

Sine the die is fair:

P(D_i)=\frac{1}{6} for i∈{1,2,3,4,5,6}

In case of 10 black and 5 white balls:

P(A|D_1)=\frac{5_{C}_1}{15_{C}_1} =\frac{5}{15}=\frac{1}{3}

P(A|D_2)=\frac{5_{C}_2}{15_{C}_2} =\frac{10}{105}=\frac{2}{21}

P(A|D_3)=\frac{5_{C}_3}{15_{C}_3} =\frac{10}{455}=\frac{2}{91}

P(A|D_4)=\frac{5_{C}_4}{15_{C}_4} =\frac{5}{1365}=\frac{1}{273}

P(A|D_5)=\frac{5_{C}_5}{15_{C}_5} =\frac{1}{3003}=\frac{1}{3003}

P(A|D_6)=\frac{5_{C}_6}{15_{C}_6} =0

Part A:

The probability that all of the balls selected are white:

P(A)=\sum^6_{i=1} P(A|D_i)P(D_i)

P(A)=\frac{1}{6}(\frac{1}{3}+\frac{2}{21}+\frac{2}{91}+\frac{1}{273}+\frac{1}{3003}+0)\\      P(A)=\frac{5}{66}=0.075757576

Part B:

The conditional probability that the die landed on 3 if all the balls selected are white:

We have to find P(D_3|A)

The data required is calculated above:

P(D_3|A)=\frac{P(A|D_3)P(D_3)}{P(A)}\\ P(D_3|A)=\frac{\frac{2}{91}*\frac{1}{6}}{\frac{5}{66} } \\P(D_3|A)=\frac{22}{455}=0.0483516

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