To solve this problem, we would use the binomial probability formula, which is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where:
- P(X = k) is the probability of having exactly k successes in n trials.
- C(n, k) is the number of combinations of n items taken k at a time, which can be calculated using the formula C(n, k) = n! / (k!(n - k)!).
- p is the probability of success on a single trial.
- n is the number of trials.
- k is the number of successes.
Given:
- The number of trials (n) is 4, since the toy is spun 4 times.
- The number of times the toy lands on the turtle (k) is 2.
The problem does not provide the actual probability of the toy landing on the turtle (p), which we need in order to calculate the desired probability. Assuming we had this probability, the following steps would be taken:
Step 1: Calculate the number of combinations where the toy lands on the turtle exactly 2 times in 4 trials using the combinations formula:
C(4, 2) = 4! / (2!(4 - 2)!)
= 4! / (2! * 2!)
= (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1))
= 24 / (2 * 2)
= 24 / 4
= 6
Step 2: Use the probability (p) of the toy landing on the turtle in each trial to calculate P(X = 2) using the binomial probability formula:
P(X = 2) = C(4, 2) * p^2 * (1 - p)^(4 - 2)
= 6 * p^2 * (1 - p)^2
Step 3: Since p is not provided, we can't complete this calculation without it. If the exact probability were given, say for example p = 0.3, then you would plug 0.3 into the place of p in the formula and compute the result:
P(X = 2) = 6 * 0.3^2 * (1 - 0.3)^2
= 6 * 0.09 * 0.49
= 6 * 0.0441
= 0.2646
Step 4: Round the result to the nearest thousandth as the problem requests. Assuming our calculated probability from the example was 0.2646:
P(X = 2) ≈ 0.265 (rounded to the nearest thousandth)
Remember, without the actual probability p, the above calculation is just an illustration. To get the real value, the exact probability of the toy landing on the turtle is required. If you provide that probability, you can use these steps to calculate the desired probability to the nearest thousandth.
