Answer:
The answer is below
Step-by-step explanation:
Let S denote syntax errors and L denote logic errors.
Given that P(S) = 36% = 0.36, P(L) = 47% = 0.47, P(S ∪ L) = 56% = 0.56
a) The probability a program contains both error types = P(S ∩ L)
The probability that the programs contains only syntax error = P(S ∩ L') = P(S ∪ L) - P(L) = 56% - 47% = 9%
The probability that the programs contains only logic error = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
P(S ∩ L) = P(S ∪ L) - [P(S ∩ L') + P(S' ∩ L)] =56% - (9% + 20%) = 56% - 29% = 27%
b) Probability a program contains neither error type= P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
c) The probability a program has logic errors, but not syntax errors = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
d) The probability a program either has no syntax errors or has no logic errors = P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
Answer:
2
Step-by-step explanation:
I think bc I just divide im guessing
Answer:
-16 is the answer after evaluating
Step-by-step explanation:
Answer:
Answer is 3/6th
Step-by-step explanation:
Because 2/4 is a half so what is 3 as a half of a fraction which is
3/6
Answer: the required probability is 0.03516
Step-by-step explanation:
Given the data in the question;
let F represent dominant allele that causes fluffy fur and
f represent corresponding recessive allele that causes wiry fur
here are the possible genotypes and corresponding phenotypes;
FF = Fluffy fur
Ff = Fluffy fur
ff = Wiry fur
genotype of parents is Ff and Ff
and there cross will be; Ff x Ff
now Offspring will be;
F f
F FF Ff
f Ff ff
so the probability that the offspring will have fluffy fur = 3/4
and the probability that the offspring will have wiry fur = 1/4
The probability that the first 2 out of 4 offspring will have fluffy fur will be;
⇒ 3/4 × 3/4 × 1/4 × 1/4 = 0.03516
Therefore, the required probability is 0.03516
