wednesday, september 24th, 2025\nbellwork non - iphones put away\n1. describe what conditional probability…

Explanation:

Step1: Identify the probabilities of making and missing

The probability of making a free - throw $p = 0.8$, and the probability of missing a free - throw $q=1 - p=0.2$. Since free - throw attempts are independent events, for independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$.

Step2: Calculate $P(\text{make }3\text{ in a row})$

Using the multiplication rule for independent events, $P(\text{make }3\text{ in a row})=0.8\times0.8\times0.8 = 0.8^{3}=0.512$.

Step3: Calculate $P(\text{make }10\text{ in a row})$

$P(\text{make }10\text{ in a row})=0.8^{10}\approx0.1074$.

Step4: Calculate $P(\text{miss }2\text{ in a row})$

$P(\text{miss }2\text{ in a row})=0.2\times0.2 = 0.2^{2}=0.04$.

Step5: Calculate $P(\text{miss }4\text{ in a row})$

$P(\text{miss }4\text{ in a row})=0.2\times0.2\times0.2\times0.2=0.2^{4}=0.0016$.

Answer for 1:

Conditional probability is the probability of an event occurring given that another event has already occurred. It measures the likelihood of an event $A$ happening under the condition that event $B$ has taken place, and it reflects how the knowledge of one event affects the probability of another.

Answer for 2:

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$, where $P(A|B)$ is the probability of event $A$ occurring given that event $B$ has occurred, $P(A\cap B)$ is the probability of both $A$ and $B$ occurring, and $P(B)$ is the probability of event $B$ occurring.

Answer for 3:

A. $P(\text{make }3\text{ in a row}) = 0.512$ B. $P(\text{make }10\text{ in a row})\approx0.1074$ C. $P(\text{miss }2\text{ in a row}) = 0.04$ D. $P(\text{miss }4\text{ in a row}) = 0.0016$