Suppose a fair four-sided die is tossed 22 times. What is the probability of getting at most 6 fours?
I did binomcdf(22,.25,6) in my calculator. n=22, p=.25, x=6
Is this the correct process?
I don't have that function on mine, so I have to actually think it through and do the math
prob(a four) = 1/4 , prob(not four) = 3/4
at most 6 fours means:
0 four, 1 four, 2 fours, 3 fours, 4 fours, 5 fours, 6 fours.
prob(your event)
= C(22,0)(3/4)^22 + C(22,1)(1/4) (3/4)^21 + ... + C(22,6) (1/4)^6 (3/4)^18
I will let you do the button-pushing
I have a feeling your binomcdf(22,.25,6)
will give you the probability of exactly 6 out of 22, but then again, I am not at all familiar with this function.
To find the probability of getting at most 6 fours when tossing a fair four-sided die 22 times, we need to consider the different scenarios that can occur.
First, we determine the total number of possible outcomes when tossing the die 22 times. Since each toss has 4 possible outcomes (1, 2, 3, or 4), the total number of possible outcomes is 4^22 (4 raised to the power of 22).
Next, we calculate the number of outcomes where we get at most 6 fours. This can happen in several ways:
- Getting exactly 0 fours: We choose 0 out of the 22 tosses to be a four, which can be done in (22 choose 0) = 1 way.
- Getting exactly 1 four: We choose 1 out of the 22 tosses to be a four, which can be done in (22 choose 1) = 22 ways.
- And so on, up to getting exactly 6 fours.
Finally, we sum up the number of outcomes for each scenario and divide it by the total number of possible outcomes to find the probability.
P(getting at most 6 fours) = (number of outcomes with 0 fours + number of outcomes with 1 four + ... + number of outcomes with 6 fours) / total number of possible outcomes
P(getting at most 6 fours) = [(22 choose 0) + (22 choose 1) + (22 choose 2) + (22 choose 3) + (22 choose 4) + (22 choose 5) + (22 choose 6)] / (4^22)
You can use a calculator or programming language to simplify the expression and compute the final probability value.