How to find probability of a and b

Answer

To find the probability of A and B (written as P(A and B) or P(A ∩ B)), multiply the probability of event A by the probability of event B. The specific formula depends on whether the events are independent or dependent.

For dependent events:

$P(A\cap B)=P(A)\times P(B|A)$

The symbol ∩ represents the intersection of two events, meaning both events happen together.

What is the multiplication rule for independent events?

The multiplication rule for independent events states that the joint probability equals the product of individual probabilities. Independent events are those where the occurrence of one event does not affect the probability of the other event.

The formula is:

$P(A\cap B)=P(A)\times P(B)$

Rolling dice provides a clear demonstration. Each roll is independent because the outcome of the first roll has no effect on the second roll.

Worked Example: Rolling Dice

Problem: What is the probability of rolling a 5 on the first die and a 2 on the second die?

Solution:

1. Identify the probability of each event.
   • P(rolling a 5) = $\frac{1}{6}$
   • P(rolling a 2) = $\frac{1}{6}$
2. Apply the multiplication rule for independent events.

$P(5\text{ and }2)=P(5)\times P(2)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$

The probability of rolling a 5 and then a 2 is $\frac{1}{36}$ or approximately 2.78%.

Worked Example: Coin Flips

Problem: What is the probability of flipping a fair coin four times and getting all heads?

Solution:

1. Each coin flip is independent with P(heads) = $\frac{1}{2}$.
2. Apply the multiplication rule for all four events.

$P(\text{4 heads})=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{16}$

The probability is $\frac{1}{16}$ or 6.25%.

What is the general multiplication rule for dependent events?

The general multiplication rule applies when events are dependent, meaning the occurrence of one event affects the probability of the other. This rule uses conditional probability.

The formula is:

$P(A\cap B)=P(A)\times P(B|A)$

The term P(B|A) represents the conditional probability of B given that A has already occurred.

This formula can be written in an equivalent form:

$P(A\cap B)=P(B)\times P(A|B)$

Both formulas yield the same result for P(A ∩ B).

Worked Example: Drawing Cards Without Replacement

Problem: Two cards are drawn from a standard 52-card deck without replacement. What is the probability that both cards are hearts?

Solution:

1. Find P(A), the probability that the first card is a heart.
   • There are 13 hearts in 52 cards.
   • P(first heart) = $\frac{13}{52}=\frac{1}{4}$
2. Find P(B|A), the probability that the second card is a heart given that the first card was a heart.
   • After drawing one heart, 12 hearts remain in 51 cards.
   • P(second heart | first heart) = $\frac{12}{51}$
3. Apply the general multiplication rule.

$P(\text{both hearts})=\frac{13}{52}\times \frac{12}{51}=\frac{156}{2652}=\frac{1}{17}$

The probability of drawing two hearts without replacement is$\frac{1}{17}$or approximately 5.88%.

Worked Example: Drawing Marbles Without Replacement

Problem: A bag contains 6 black marbles and 4 blue marbles. Two marbles are drawn without replacement. What is the probability that both marbles are blue?

Solution:

1. Find P(A), the probability of drawing a blue marble first.
   • P(first blue) = $\frac{4}{10}=\frac{2}{5}$
2. Find P(B|A), the probability of drawing a blue marble second given the first was blue.
   • After removing one blue marble, 3 blue marbles remain in 9 total marbles.
   • P(second blue | first blue) = $\frac{3}{9}=\frac{1}{3}$
3. Apply the general multiplication rule.

$P(\text{both blue})=\frac{4}{10}\times \frac{3}{9}=\frac{12}{90}=\frac{2}{15}$

The probability of drawing two blue marbles is $\frac{2}{15}$ or approximately 13.33%.

What is conditional probability?

Conditional probability measures the probability of an event occurring given that another event has already occurred. The notation P(A|B) reads as "the probability of A given B."

The formula for conditional probability is:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

This formula requires P(B) > 0.

Rearranging this formula produces the general multiplication rule:

$P(A\cap B)=P(B)\times P(A|B)$$P(A\cap B)=P(B)\times P(A|B)$

For independent events, P(A|B) = P(A) because the occurrence of B does not change the probability of A. Substituting this into the general multiplication rule gives the simplified formula $P(A\cap B)=P(A)\times P(B)$.

How do you extend the multiplication rule to three or more events?

The multiplication rule extends to any number of events by incorporating successive conditional probabilities.

For three events A, B, and C:

$P(A\cap B\cap C)=P(A)\times P(B|A)\times P(C|A\cap B)$

For n events A₁, A₂, ..., Aₙ:

$P(A_1\cap A_2\cap \dots \cap A_n)=P(A_1)\times P(A_2|A_1)\times P(A_3|A_1\cap A_2)\times \dots \times P(A_n|A_1\cap A_2\cap \dots \cap A_{n-1})$

For n independent events, this simplifies to:

$P(A_1\cap A_2\cap \dots \cap A_n)=P(A_1)\times P(A_2)\times \dots \times P(A_n)$

Worked Example: College Acceptance

Problem: A student has a 0.80 probability of college acceptance. Of accepted students, 60% receive dormitory housing. Of those with dormitory housing, 80% have at least one roommate. What is the probability of being accepted, receiving dormitory housing, and having no roommates?

Solution:

1. Identify the probabilities.
   • P(Accepted) = 0.80
   • P(Dormitory | Accepted) = 0.60
   • P(No Roommates | Dormitory and Accepted) = 1 - 0.80 = 0.20
2. Apply the extended multiplication rule.

$P(\text{Accepted and Dormitory and No Roommates})=0.80\times 0.60\times 0.20=0.096$

The probability is 0.096 or 9.6%.$P(A_1\cap A_2\cap \dots \cap A_n)=P(A_1)\times P(A_2)\times \dots \times P(A_n)$$\frac{\mathrm{<span\; style="font-size:\; 117.2\%;\; font-family:\; sans-serif;\; letter-spacing:\; 0.01rem;">17</span>}}{}$

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