Answer:
probability that all of the sprinklers will operate correctly in a fire: 0.0282
Step-by-step explanation:
In order to solve this question we will use Binomial probability distribution because:
- In the question it is given that the sprinklers activate correctly or not independently.
- The number of outcomes are two i.e. sprinklers activate correctly or not.
A binomial distribution is a probability of a success or failures outcomes in an repeated multiple or n times.
Number of outcomes of this distributions are two.
The formula is:
b(x; n, P) = 
b = binomial probability also represented as P(X=x)
x =no of successes
P = probability of a success on a single trial
n = no of trials
is calculated as:
= n! / x!(n – x)!
= 10! / 10!(10-10)!
= 1
According to given question:
probability of success i.e. p = 0.7 i.e. probability of a sprinkler to activate correctly.
number of trials i.e. n = 10 as number of sprinklers are 10
To find: probability that all of the sprinklers will operate correctly in a fire
X = 10 because we have to find the probability that "all" of the sprinklers will operate correctly and there are 10 sprinklers so all 10 of them
So putting these into the formula:
P(X=x) = 
= C₁₀,₁₀ * 0.7¹⁰ * (1-0.7)¹⁰⁻¹⁰
= 1 * 0.0282 * (0.3) ⁰
= 1 * 0.0282 * 1
P(X=x) = 0.0282
Answer: 720ways, 24ways
Step-by-step explanation:
Given the seven letter words "SYSTEMS", if E is always occurring before M it means E and M will always be together therefore they letter 'EM' will be taken as an entity to five us 6letters i.e SYST(EM)S.
This can then be arranged in 6!ways
6! = 6×5×4×3×2×1 = 720ways
Similarly, if the E somewhere before the M and the three Ss grouped consecutively, this means E and M must always be together as well as the Ss to give (SSS)YT(EM).
This means that the letters in the bracket can be taken as an entity to give a total of 4 entities. This can them be arranged in 4! ways.
4! = 4×3×2×1
4! = 24ways
2 1/2 x 3 = 7 1/2 = 7 4/8
10 11/8
- 7 4/8
3 7/8
A is the correct answer
Answer:
Step-by-step explanation:
2 * 2 = 4
3*5 = 15
2*5 = 10
4+15+10 = 29 