7) Suppose that you roll a pair of standard 6-sided dice a total of 42 times. What is the probability that you will get a sum of 5 at least four times?
The probability of getting 5 on one toss of a pair is 1/9. The most probable number of fives you will get after 42 tosses is 4 or 5.
Add the probabilities of getting 5 four, five, six, seven, eight etc timesout of 42 tosses. By the time you get to 15 times, the probability will be very small and you can quit adding them up.
The probability of rolling 5 four times is (1/9)^4*(8/9)^38*42!/(38!*4!)
= 119,930*1.524*10^-4*1.138*10^-2
= 0.20803
The probability of rolling 5 five times is (1/9)^5*(8/9)^37*42!/(37!*5!)
= 8.507*10^5*1.6935*10^-5*1.2804*10^-2
= 0.18447
I am not going to continue this because it would take half the night, but you get the idea.
There is an easier way to do this that just occurred to me. It will give the same answer. Here it is:
Add the probabilities of getting 5 zero, one, two or three times, and subtract that from 1.
The probability of never getting five is (8/9)^42 = 0.007105
The probability of getting 5 only once is (1/9)*(8/9)^41*42 = 0.037304
The probability of getting 5 twice is
(1/9)^2*(8/9)^40*(42*41*/2) = 0.095591
Now YOU compute the probability of getting 5 three times. Add that to the probabilites above and subtract the sum from one. You should end up with a number around 0.6.
That will be the answer.
To calculate the probability of getting a sum of 5 at least four times, we will need to find the probability of getting a sum of 5 exactly four times, exactly five times, six times, and so on, up to 42 times.
Let's break down the problem step-by-step:
Step 1: Calculate the probability of getting a sum of 5 on a single roll:
To get a sum of 5, there are four possible combinations: (1, 4), (2, 3), (3, 2), and (4, 1). Since there are a total of 36 equally likely outcomes (each die has 6 sides), the probability of getting a sum of 5 on a single roll is 4/36 = 1/9.
Step 2: Calculate the probability of getting a sum of 5 exactly four times:
To calculate this, we use the binomial probability formula. The formula is given by:
P(X = k) = nCk * 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
k is the number of successes
p is the probability of getting a success on a single trial
In our case,
n = 42 (total number of times the dice are rolled)
k = 4 (number of times we want to get a sum of 5)
p = 1/9 (probability of getting a sum of 5 on a single roll)
Using the formula, the probability of getting a sum of 5 exactly four times is:
P(X = 4) = 42C4 * (1/9)^4 * (8/9)^(42-4)
Step 3: Calculate the probability of getting a sum of 5 exactly five times:
Using the same formula as in Step 2, with k = 5, we have:
P(X = 5) = 42C5 * (1/9)^5 * (8/9)^(42-5)
Repeat this calculation for k = 6, 7, ..., up to 42.
Step 4: Calculate the probability of getting a sum of 5 at least four times:
To find this, we need to add up the probabilities from Step 2 to Step 3:
P(at least 4 times) = P(X = 4) + P(X = 5) + P(X = 6) + ... + P(X = 42)
This will give us the final probability that you will get a sum of 5 at least four times.
Hope this helps!
To find the probability of getting a sum of 5 at least four times when rolling a pair of standard 6-sided dice 42 times, we need to break down the problem into smaller steps.
Step 1: Find the probability of rolling a sum of 5 on a single roll
When rolling two standard 6-sided dice, there are multiple ways to get a sum of 5: (1, 4), (2, 3), (3, 2), (4, 1). Out of the 36 possible outcomes (6 possibilities for the first die and 6 possibilities for the second die), there are 4 favorable outcomes to get a sum of 5. Therefore, the probability of rolling a sum of 5 on a single roll is 4/36, which simplifies to 1/9.
Step 2: Calculate the probability of getting a sum of 5 less than four times
To find the probability of getting a sum of 5 less than four times, we need to consider the complementary event. This means we calculate the probability of not getting a sum of 5 four or more times.
The number of ways to choose four rolls out of 42 to be a sum of 5 is given by the binomial coefficient `(42 choose 4)`, which is calculated as:
C(42, 4) = 42! / (4! * (42-4)!) = 42! / (4! * 38!)
So, the probability of getting a sum of 5 exactly four times is [(42 choose 4) * (1/9)^4 * (8/9)^(42-4)]. However, we're interested in the probability of getting a sum of 5 at least four times.
Step 3: Calculate the probability of getting a sum of 5 at least four times
To find the probability of getting a sum of 5 at least four times, we need to consider the complement of this event. This means we calculate the probability of not getting a sum of 5 three or fewer times.
The number of ways to choose three rolls out of 42 to be a sum of 5 is given by the binomial coefficient `(42 choose 3)`, which is calculated as:
C(42, 3) = 42! / (3! * (42-3)!) = 42! / (3! * 39!)
So, the probability of not getting a sum of 5 exactly three or fewer times is [(42 choose 3) * (1/9)^3 * (8/9)^(42-3)].
Step 4: Find the desired probability
Now that we have the probability of not getting a sum of 5 three or fewer times, we can find the probability of getting a sum of 5 at least four times by subtracting this probability from 1.
Probability of getting a sum of 5 at least four times = 1 - [ (42 choose 3) * (1/9)^3 * (8/9)^(42-3) ]
By evaluating this expression, you can find the probability of getting a sum of 5 at least four times when rolling a pair of standard 6-sided dice 42 times.