Sure, let’s solve each part of this problem step by step:
1. Probability that a child is NOT allergic to nuts:
The probability of not being allergic to nuts (¬nuts) = 1 - Probability of being allergic to nuts
Probability of being allergic to nuts = 0.014
Therefore, Probability of not being allergic to nuts = 1 - 0.014 = 0.986
2. Probability that a child is NOT allergic to dairy and NOT allergic to gluten:
Probability of not being allergic to dairy (¬dairy) = 1 - Probability of being allergic to dairy = 1 - 0.043 = 0.957
Probability of not being allergic to gluten (¬gluten) = 1 - Probability of being allergic to gluten = 1 - 0.031 = 0.969
Probability of not being allergic to dairy and not being allergic to gluten = Probability of not being allergic to dairy * Probability of not being allergic to gluten = 0.957 * 0.969 = 0.927
3. Probability that a child is allergic to nuts or is allergic to dairy or is allergic to gluten:
P(allergic to nuts) = 0.014
P(allergic to dairy) = 0.043
P(allergic to gluten) = 0.031
Using the inclusion-exclusion principle for probabilities, the formula is:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)
Here, A represents the event of being allergic to nuts, B for dairy, and C for gluten. As each event is independent, the intersection of any two events would be negligible.
Therefore, P(allergic to nuts or allergic to dairy or allergic to gluten) = 0.014 + 0.043 + 0.031 - 0 - 0 - 0 + 0 = 0.088
4. Probability that a child is allergic to dairy and is allergic to gluten:
P(allergic to dairy and allergic to gluten) = P(allergic to dairy) * P(allergic to gluten) = 0.043 * 0.031 = 0.001333
5. Probability that in a room full of 6 children, NO child is allergic to nuts:
Probability of no child being allergic to nuts = Probability of not being allergic to nuts for one child ^ Number of children
Probability of no child being allergic to nuts = 0.986 ^ 6 = 0.92537492222