To solve the problem of finding the probability that an event happens at least three times in five days, given that the probability of the event happening on any day is 0.2, we can model this as a binomial distribution.

The binomial distribution formula is: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where:

• $n$ is the number of trials (days in this case, $n = 5$),
• $k$ is the number of successes (the event happening),
• $p$ is the probability of success on each trial (0.2 here),
• $\binom{n}{k}$ is the binomial coefficient, which gives the number of ways to choose $k$ successes from $n$ trials.

We need to find the probability that the event happens at least 3 times, i.e., $P(X \geq 3)$. This can be written as: $P(X \geq 3) = 1 - P(X < 3)$ We calculate $P(X < 3)$, which is the sum of probabilities for 0, 1, or 2 successes: $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ Now let's compute these probabilities step by step.

1. $P(X = 0)$: $P(X = 0) = \binom{5}{0} (0.2)^0 (0.8)^5 = 1 \times 1 \times (0.8)^5 = 0.32768$
2. $P(X = 1)$: $P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^4 = 5 \times 0.2 \times (0.8)^4 = 5 \times 0.2 \times 0.4096 = 0.4096$
3. $P(X = 2)$: $P(X = 2) = \binom{5}{2} (0.2)^2 (0.8)^3 = 10 \times (0.2)^2 \times (0.8)^3 = 10 \times 0.04 \times 0.512 = 0.2048$

Now we can sum these probabilities: $P(X < 3) = 0.32768 + 0.4096 + 0.2048 = 0.94208$

Finally, we find $P(X \geq 3)$: $P(X \geq 3) = 1 - P(X < 3) = 1 - 0.94208 = 0.05792$

Thus, the probability that the event happens at least three times in five days is approximately 0.0579 or 5.79%.

Would you like further details or have any questions?

Related Questions:

1. How would the probability change if the event's daily probability were increased to 0.3?
2. What is the probability of the event happening exactly four times in five days?
3. How would you calculate the expected number of days the event occurs in this scenario?
4. Can you generalize this problem to $n$ days with a probability $p$?
5. How would the solution differ if the event’s occurrence on each day was dependent on previous days?

Tip:

In binomial probability, when calculating probabilities for "at least" or "at most" scenarios, it's often easier to use the complement rule.