Solveeit Logo
Login

Question

Question: If ‘A’ and ‘B’ two events such that P (C) \(={\dfrac{1}{4}}.\) P (B) \(=\dfrac{1}{8},\) the find P (...

If ‘A’ and ‘B’ two events such that P (C) =14.={\dfrac{1}{4}}. P (B) =18,=\dfrac{1}{8}, the find P (not A and not B)

Explanation

Solution

Remember the formula P (A or B)=P(A)+P(B)− P (A∩ B)\left( \text{A}\,\text{or}\ \text{B} \right)=\text{P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\,\text{P}\,\left( \text{A}\cap \,\text{B} \right) for P (not A and not B) subtract P(A and B) from 1 because we know that the sum of probability of a work to happen and not happen is always 1.

Complete step by step solution:
Given, that these are 2 events A and P also
Probability of A = P (A)
P (A) = 14\dfrac{1}{4} (given)
Probability of B = P (B)
P(B) = P12\text{P}\dfrac{1}{2} (given)
Also
P (A ∩ B)18\text{P}\,\left( \text{A}\,\cap \,\text{B} \right)\dfrac{1}{8}
We have to find P (not A and not B)
P (A !!′!! ∩ B !!′!! )\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right) = we have to find . Now,
There is a formula for P (A !!′!! ∩ B !!′!! )\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right) not is can also be rotten like P (A !!′!! ∩ B !!′!! )\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right)

P (A !!′!!  ∩ B !!′!! ) = P (A∪B)\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\,\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right)\,=\,\text{P}\,\left( \text{A}\cup \text{B} \right)
If we solve now, we get
P (not A not B) = P (A !!′!! ∩ B !!′!! )\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right)
= P (A !!′!! ∩ B !!′!! )\text{P}\,\left( \text{A }\\!\\!'\\!\\!\text{ }\cap \,\text{B }\\!\\!'\\!\\!\text{ } \right)
1-P (A or B)\text{1-P}\,\left( \text{A}\,\text{or}\,\text{B} \right)
[Therefore, P (A or B) = P(A)+P(B)−P(A∩B)] = 1−[P (A)+P(B)−P(A∩B)][Therefore, \,\text{P}\,\left( \text{A}\,\text{or}\,\text{B} \right)\,=\,\text{P}(\text{A})+\text{P}(\text{B})-P(\text{A}\cap \text{B})]\,=\,1-[\text{P}\,(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap \text{B})]
= 1−(14+ 12− 18)1-\left( {\dfrac{1}{4}}+\ \dfrac{1}{2}-\ \dfrac{1}{8} \right)
=1−(28+48−18)=1-\left( \dfrac{2}{8}+\dfrac{4}{8}-\dfrac{1}{8} \right)
=1−58=1-\dfrac{5}{8}
=38=\dfrac{3}{8}

The value for P (not A not B) =38=\dfrac{3}{8}

Note: In this type of question first remember the formula and then try to find the suitable formula according to the given information from the question and also we need to know the representation of everything like in this question
P(not A not B) = P (A’∩B’) = P ( A ∪B’)
This is also equally important.