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Two events are independent when the outcome of one event does not affect the outcome of the second event.
There are two important formulas for calculating the probability that independent events occur.
Independent Events Probability Formulas
Probability of A and B
\( P\left( A \text{ and } B \right) = P(A) \cdot P(B) \)
This is the probability that both events occur.
Probability of A or B
\( P\left( A \text{ or } B \right) = P\left( A \right) + P\left( B \right) - P\left( A \text{ and } B \right) \)
This is the probability that at least one of the events occurs.
Note that you can find the probability of more than two independent events occurring. This is beyond the scope of this lesson, but it is important to understand that independent events do not have to occur in pairs.
Give It a Shot!
How well can you use the formulas above to find the probabilities of independent events? Read the scenario on each tab. Then calculate the probability. Be sure to check your answers.
What is the probability of landing a fair coin on tails and rolling a 6 on a fair six-sided die?
The probability is 0.0833 or 8.33%.
If you need help arriving at this answer, click the Solution button.
| From the question, you know this is an AND probability. |
| Use the formula \( P\left( A \text{ and } B \right) = P\left( A \right) \cdot P\left( B \right) \) |
| Define the events. |
| Event A = “landing on tails” and event B = ‘rolling a 6. |
| Determine the probability of each event. |
The probability of landing on tails is P(A) = \( \frac{1}{2} \). The probability of rolling a 6 is P(B) = \( \frac{1}{6} \). |
| Use the formula. |
Substitute. \( P\left( A \text{ and } B \right) = \frac{1}{2} \cdot \frac{1}{6} \) \( P\left( A \text{ and } B \right) = \frac{1}{12} = 0.0833 = 8.33\% \). |
A bag contains 40 colored pencils. There are 10 are green pencils, 5 grey, 12 yellow, and 13 red.
What is the probability of randomly selecting a red pencil out of the bag, replacing it, and then selecting a green pencil out of the bag?
The probability is 0.0813 or 8.13%.
If you need help arriving at this answer, click the Solution button.
| From the question, you know this is an AND probability. |
| Use the formula \( P\left( A \text{ and } B \right) = P\left( A \right) \cdot P\left( B \right) \) |
| Define the events. |
| Let event A = "select a red pencil" and event B = "select a green pencil." |
| Determine the probability of each event. |
| The probability of selecting a red pencil is P(A) = \( \frac{13}{40} \).
The probability of selecting a green pencil is P(B) = \( \frac{10}{40} \) = \( \frac{1}{4} \) Simplifying the fraction for P(B) now will ease calculations later. |
| Use the formula. |
| Substitute.
\( P\left( A \text{ and } B \right) = \frac{13}{40} \cdot \frac{1}{4} \) \( P\left( A \text{ and } B \right) = \frac{13}{160} = 0.08125 = 8.125\% \) |
A carnival has a spin-the-wheel game and a ring-toss game. The probability of winning the spin-the-wheel game is 15%, and the probability of winning the ring-toss game is 10%.
What is the probability of winning one of the games?
The probability is \( 0.235 \text{ or } 23.5\% \)
If you need help arriving at this answer, click the Solution button.
| From the question, you know this is an AND probability. |
| Use the formula \( P\left( A \text{ or } B \right) = P\left( A \right) + P\left( B \right) - P\left( A \text{ and } B \right) \) |
| Define the events. |
| Let event A = "winning spin-the-wheel" and let event B = "winning ring-toss." |
| Determine the probability of each event. |
| Express your probabilities as decimals to help with calculation. P(A) = 0.15 and P(B) = 0.10. |
| To use the OR probability formula, you need to calculate \( \left( A \text{ and } B \right) = P\left( A \right) \cdot P\left( B \right) \). |
| Substitute. \( P\left( A \text{ and } B \right) = \left( 0.15 \right)\left( 0.10 \right) \) \( P\left( A \text{ and } B \right) = 0.015 \) |
| Use the OR formula. |
| Substitute. \( P\left( A \text{ or } B \right) = \left( 0.15 \right) + \left( 0.10 \right) - \left( 0.015 \right) \) \( P\left( A \text{ or } B \right) = 0.235 = 23.5\% \) |
Two bags are filled with candy. Bag 1 has 20 chocolate bars and 30 sour candy treats. Bag 2 has 15 chocolate bars, 15 sour candy treats, and 10 pieces of sweet candy.
What is the probability of selecting a chocolate bar from Bag 1 or selecting a sour candy treat from Bag 2?
The probability is \( \frac{5}{8} \text{ or } 62.5\%. \)
If you need help arriving at this answer, click the Solution button.
| From the question, you know this is an AND probability. |
| Use the formula \( P\left( A \text{ or } B \right) = P\left( A \right) + P\left( B \right) - P\left( A \text{ and } B \right) \) |
| Define the events. |
| Let A = "selecting a chocolate bar from Bag 1" and B = "selecting a sour candy treat from Bag 2." |
| Determine the probability of each event. |
| Express your probabilities as decimals to help with calculation. \( P(A) = \frac{20}{50} = 0.40 \) \( P(B) = \frac{15}{40} = 0.375 \) |
| Calculate \( P\left( A \text{ and } B \right) = P\left( A \right) \cdot P\left( B \right) \). |
| Substitute. \( P\left( A \text{ and } B \right) = (0.40)(0.375) \) \( P\left( A \text{ and } B \right) = 0.15 \) |
| Use the OR formula. |
| Substitute. \( P\left( A \text{ or } B \right) = \left( 0.40 \right) + \left( 0.375 \right) - (0.15) \) \( P\left( A \text{ or } B \right) = 0.625 = 62.5\% \) |
