A person decides to tease a neighbor's dog. The dog bites in 30% of cases when people tease it. The person decides to tease the dog no more than 3 times, up to the first bite. $X$ is the number of attempts to tease the dog in which the dog does not bite. We need to create a probability distribution for the random variable $X$.

1. Problem Description

A person decides to tease a neighbor's dog. The dog bites in 30% of cases when people tease it. The person decides to tease the dog no more than 3 times, up to the first bite. $X$ is the number of attempts to tease the dog in which the dog does not bite. We need to create a probability distribution for the random variable $X$.

2. Solution Steps

Let $p$ be the probability of the dog biting, so $p = 0.3$.

Let $q$ be the probability of the dog not biting, so $q = 1 - p = 1 - 0.3 = 0.7$.

The random variable $X$ can take values 0, 1,

2. If $X=0$, it means the dog bites on the first try. The probability is $P(X=0) = p = 0.3$.

If $X=1$, it means the dog doesn't bite on the first try and bites on the second try. The probability is $P(X=1) = q \cdot p = 0.7 \cdot 0.3 = 0.21$.

If $X=2$, it means the dog doesn't bite on the first and second tries, and bites on the third try. The probability is $P(X=2) = q \cdot q \cdot p = q^2 \cdot p = (0.7)^2 \cdot 0.3 = 0.49 \cdot 0.3 = 0.147$.

The probability distribution is as follows:

$P(X=0) = 0.3$

$P(X=1) = 0.21$

$P(X=2) = 0.147$

Let's verify that the probabilities sum up to less than or equal to

1. $0.3 + 0.21 + 0.147 = 0.657$. The person stops after the first bite or after 3 tries.

The complement probability is $1 - 0.657 = 0.343$. This probability corresponds to that the dog did not bite after the 3 tries and the person just stopped trying to tease it. However, it is given that the variable $X$ represent the number of attempts to tease the dog in which the dog does *not* bite.

3. Final Answer

The probability distribution of $X$ is:

$P(X=0) = 0.3$

$P(X=1) = 0.21$

$P(X=2) = 0.147$