Suppose you flip an unfair coin, where 𝑝(ℎ𝑒𝑎𝑑𝑠)=0.65.

A) What is the probability of getting AT MOST 7 heads on 10 flips?
B) What is the probability of getting at least 6 heads on 10 flips?

It's the "At most" that is messing me up. Anyone?

at most 7 heads is the same as at least 3 tails, right?

P(tails) = 0.35

Omg.... (face smack). Thank you.

To solve these problems, we can use the binomial probability formula. The binomial probability formula is given by:

P(X=k) = (n C k) * p^k * (1-p)^(n-k)

where:
- P(X=k) is the probability of getting exactly k successes,
- n is the total number of trials,
- p is the probability of success in a single trial,
- (n C k) is the number of combinations of n things taken k at a time,
- k is the number of successes.

Now let's solve each part:

A) To find the probability of getting AT MOST 7 heads on 10 flips:

P(X ≤ 7) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=7)

We can calculate each individual probability using the binomial probability formula and sum them up.

B) To find the probability of getting at least 6 heads on 10 flips:

P(X ≥ 6) = P(X=6) + P(X=7) + P(X=8) + ... + P(X=10)

Again, we can calculate each individual probability using the binomial probability formula and sum them up.

Let's calculate them step-by-step:

A) Probability of getting AT MOST 7 heads on 10 flips:

P(X=0) = (10 C 0) * (0.65^0) * (0.35^10)
P(X=1) = (10 C 1) * (0.65^1) * (0.35^9)
P(X=2) = (10 C 2) * (0.65^2) * (0.35^8)
...
P(X=7) = (10 C 7) * (0.65^7) * (0.35^3)

Then, sum up all these probabilities.

B) Probability of getting at least 6 heads on 10 flips:

P(X=6) = (10 C 6) * (0.65^6) * (0.35^4)
P(X=7) = (10 C 7) * (0.65^7) * (0.35^3)
P(X=8) = (10 C 8) * (0.65^8) * (0.35^2)
...
P(X=10) = (10 C 10) * (0.65^10) * (0.35^0)

Then, sum up all these probabilities.

I hope this clears things up! Let me know if you need further assistance.

To calculate the probability of getting at most 7 heads on 10 flips with an unfair coin, you can use the binomial probability formula. Let's break it down step by step:

A) Probability of getting exactly 𝑘 heads on 𝑛 flips with probability of heads (𝑝) and tails (1-𝑝):
𝑃(𝑘 heads) = 𝑛C𝑘 * 𝑝^𝑘 * (1-𝑝)^(𝑛-𝑘)

The probability of getting at most 7 heads on 10 flips can be calculated by adding up the probabilities of getting 0, 1, 2, 3, 4, 5, 6, and 7 heads:

𝑃(𝑎𝑡 𝑚𝑜𝑠𝑡 7 heads) = 𝑃(𝑘=0) + 𝑃(𝑘=1) + 𝑃(𝑘=2) + 𝑃(𝑘=3) + 𝑃(𝑘=4) + 𝑃(𝑘=5) + 𝑃(𝑘=6) + 𝑃(𝑘=7)

To calculate this, you need to substitute 𝑝 = 0.65, 𝑛 = 10, and calculate each individual term using the binomial probability formula mentioned earlier. Then sum up the results.

B) Similarly, to calculate the probability of getting at least 6 heads on 10 flips, you can use the same approach. You need to calculate the probabilities of getting 6, 7, 8, 9, and 10 heads and then sum them up:

𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 6 heads) = 𝑃(𝑘=6) + 𝑃(𝑘=7) + 𝑃(𝑘=8) + 𝑃(𝑘=9) + 𝑃(𝑘=10)

Again, substitute 𝑝 = 0.65, 𝑛 = 10, and use the binomial probability formula to calculate each term before summing them up.

To make the calculations simpler, you can use a graphing calculator or an online binomial probability calculator that can directly provide you with the results for a given probability, number of trials, and desired outcomes.