A person decides to tease a neighbor's dog. The neighbor says that 30% of people teasing the dog get bitten. The person decides to try teasing the dog at most 3 times, until the first bite. $X$ is the number of attempts to tease the dog in which the dog does not bite. Create the distribution series and plot the distribution polygon of the random variable $X$.

1. Problem Description

A person decides to tease a neighbor's dog. The neighbor says that 30% of people teasing the dog get bitten. The person decides to try teasing the dog at most 3 times, until the first bite. $X$ is the number of attempts to tease the dog in which the dog does not bite. Create the distribution series and plot the distribution polygon of the random variable $X$.

2. Solution Steps

Let $p$ be the probability of being bitten, which is $p = 0.3$. Then the probability of not being bitten is $q = 1 - p = 1 - 0.3 = 0.7$.

$X$ can take the values 0, 1,

2.

If $X = 0$, the person is bitten on the first attempt. The probability is $P(X = 0) = p = 0.3$.

If $X = 1$, the person is not bitten on the first attempt, and bitten on the second attempt. The probability is $P(X = 1) = q \cdot p = 0.7 \cdot 0.3 = 0.21$.

If $X = 2$, the person is not bitten on the first and second attempts, and bitten on the third attempt. 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 distribution series is:

$X$: 0, 1, 2

$P(X)$: 0.3, 0.21, 0.147

The sum of probabilities is $0.3 + 0.21 + 0.147 = 0.657$.

Since the person attempts at most 3 times until the first bite, it is important to ensure that the probabilities sum up to

1. However, the given probabilities do not sum up to

1. This means that there is a probability that the person is not bitten in any of the three trials.

The probability that the person is not bitten in any of the three attempts is $P(X > 2) = q^3 = (0.7)^3 = 0.343$.

Therefore, the modified probabilities accounting for at most 3 attempts are:

$P(X=0) = 0.3$

$P(X=1) = 0.7 * 0.3 = 0.21$

$P(X=2) = 0.7 * 0.7 * 0.3 = 0.147$

$P(\text{no bite in 3 attempts}) = 0.7 * 0.7 * 0.7 = 0.343$

Revised probabilities:

$X$: 0, 1, 2

$P(X)$: 0.3, 0.21, 0.147

Let's verify the sum: 0.3 + 0.21 + 0.147 + 0.343 =

1.

The distribution polygon (histogram) has vertices (0, 0.3), (1, 0.21), (2, 0.147).

Note: The last case would not be plotted with the polygon itself.

3. Final Answer

Distribution series:

$X$: 0, 1, 2

$P(X)$: 0.3, 0.21, 0.147