\(Evaluate \ P(A∩B)\ if \ 2P(A) = P(B) =\) \(\frac {5}{13}\) \(and \ P(A|B)=\) \(\frac 25\)
\(Given:\)
\(2P(A) = P(B) =\) \(\frac {5}{13}\),
\(P(A|B)=\frac 25\)
\(∴P(A)=\frac {5}{26}\)
\(Now,\) \(P(A∩B)=P(B|A).P(B)\)
\(P(A∩B)=\frac 25\times\frac {5}{13}\)
\(P(A∩B) =\frac {2}{13}\)
\(Now,\) \(P(A∪B)=P(A)+P(B)-P(A∩B)\)
\(P(A∪B) =\frac {5}{26}+\frac {5}{13}-\frac {2}{13}\)
\(P(A∪B) =\frac {11}{26}\)
A shop selling electronic items sells smartphones of only three reputed companies A, B, and C because chances of their manufacturing a defective smartphone are only 5%, 4%, and 2% respectively. In his inventory, he has 25% smartphones from company A, 35% smartphones from company B, and 40% smartphones from company C.
A person buys a smartphone from this shop
A shop selling electronic items sells smartphones of only three reputed companies A, B, and C because chances of their manufacturing a defective smartphone are only 5%, 4%, and 2% respectively. In his inventory, he has 25% smartphones from company A, 35% smartphones from company B, and 40% smartphones from company C.
A person buys a smartphone from this shop
(i) Find the probability that it was defective.
Commodities | 2009-10 | 2010-11 | 2015-16 | 2016-17 |
---|---|---|---|---|
Agriculture and allied products | 10.0 | 9.9 | 12.6 | 12.3 |
Ore and minerals | 4.9 | 4.0 | 1.6 | 1.9 |
Manufactured goods | 67.4 | 68.0 | 72.9 | 73.6 |
Crude and petroleum products | 16.2 | 16.8 | 11.9 | 11.7 |
Other commodities | 1.5 | 1.2 | 1.1 | 0.5 |
Categories of Reporting Area | As a percentage of total cultivable land (1950-51) | As a percentage of total cultivable land (2014-15) | Area (1950-51) | Area (2014-15) |
---|---|---|---|---|
Culturable waste land | 8.0 | 4.0 | 13.4 | 6.8 |
Fallow other than current fallow | 6.1 | 3.6 | 10.2 | 6.2 |
Current fallow | 3.7 | 4.9 | 6.2 | 8.4 |
Net area sown | 41.7 | 45.5 | 70.0 | 78.4 |
Total Cultivable Land | 59.5 | 58.0 | 100.00 | 100.00 |
Conditional Probability is defined as the occurrence of any event which determines the probability of happening of the other events. Let us imagine a situation, a company allows two days’ holidays in a week apart from Sunday. If Saturday is considered as a holiday, then what would be the probability of Tuesday being considered a holiday as well? To find this out, we use the term Conditional Probability.
P(S | B) = P(B | B) = 1.
Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.
[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].
P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.
This theorem is named as the Multiplication Theorem of Probability.
Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.
We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).
So, P(A ∩ B) = P(A). P(B | A).
Similarly, P(A ∩ B) = P(B). P(A | B).
The interesting information regarding the Multiplication Theorem is that it can further be extended to more than two events and not just limited to the two events. So, one can also use this theorem to find out the conditional probability in terms of A, B, or C.
Read More: Types of Sets
Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.