Suppose you flip an unfair coin, where π(βππππ )=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 β€ 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 β₯ 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 π heads on π flips with probability of heads (π) and tails (1-π):
π(π heads) = πCπ * π^π * (1-π)^(π-π)
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:
π(ππ‘ πππ π‘ 7 heads) = π(π=0) + π(π=1) + π(π=2) + π(π=3) + π(π=4) + π(π=5) + π(π=6) + π(π=7)
To calculate this, you need to substitute π = 0.65, π = 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:
π(ππ‘ ππππ π‘ 6 heads) = π(π=6) + π(π=7) + π(π=8) + π(π=9) + π(π=10)
Again, substitute π = 0.65, π = 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.