To construct a binomial distribution with $N = 6$ (number of trials) and $P = 0.32$ (probability of success for each trial), we can calculate the probability of obtaining $k$ successes out of $N$ trials using the binomial probability formula:

$P(X = k) = \binom{N}{k} P^k (1 - P)^{N-k}$

Where:

• $P(X = k)$ is the probability of getting $k$ successes,
• $\binom{N}{k} = \frac{N!}{k!(N-k)!}$ is the binomial coefficient,
• $N = 6$,
• $P = 0.32$, and
• $1 - P = 0.68$ (probability of failure).

We will compute the probabilities for $k = 0, 1, 2, 3, 4, 5, 6$.

Step-by-step calculation:

1. For $k = 0$: $P(X = 0) = \binom{6}{0} (0.32)^0 (0.68)^6 = 1 \times 1 \times 0.68^6 = 0.0886$

2. For $k = 1$: $P(X = 1) = \binom{6}{1} (0.32)^1 (0.68)^5 = 6 \times 0.32 \times 0.68^5 = 0.2498$

3. For $k = 2$: $P(X = 2) = \binom{6}{2} (0.32)^2 (0.68)^4 = 15 \times 0.32^2 \times 0.68^4 = 0.3313$

4. For $k = 3$: $P(X = 3) = \binom{6}{3} (0.32)^3 (0.68)^3 = 20 \times 0.32^3 \times 0.68^3 = 0.2226$

5. For $k = 4$: $P(X = 4) = \binom{6}{4} (0.32)^4 (0.68)^2 = 15 \times 0.32^4 \times 0.68^2 = 0.0886$

6. For $k = 5$: $P(X = 5) = \binom{6}{5} (0.32)^5 (0.68)^1 = 6 \times 0.32^5 \times 0.68 = 0.0171$

7. For $k = 6$: $P(X = 6) = \binom{6}{6} (0.32)^6 (0.68)^0 = 1 \times 0.32^6 \times 1 = 0.0011$

Summary of the binomial distribution:

$k$ (Number of successes)	Probability $P(X = k)$
0	0.0886
1	0.2498
2	0.3313
3	0.2226
4	0.0886
5	0.0171
6	0.0011

This table provides the probabilities for each possible number of successes when performing 6 trials, with a success probability of 0.32.

Would you like more details or have any questions?

Related questions:

1. How would the distribution change if $P$ were 0.5 instead of 0.32?
2. What is the expected value (mean) and variance of this binomial distribution?
3. How can we plot the binomial distribution for $N = 6$ and $P = 0.32$?
4. What is the probability of getting at most 2 successes?
5. How does the shape of the binomial distribution change as $N$ increases?

Tip:

In a binomial distribution, the mean is $\mu = N \times P$ and the variance is $\sigma^2 = N \times P \times (1 - P)$.