Conditioning restricts the sample space. For P(B)>0, P(A B)= P(A B)P(B), so P(A B)=P(B)P(A B). Divide by the probability of the given event, not the event being asked about. Example: a bag contains 3 red and 2 blue balls. Without replacement, P(two red)= 35 x 24= 310. With replacement and independent draws, it is 35 x 35= 925. To test independence compare P(A B) with P(A)P(B), or compare P(A B) with P(A). If the relevant probabilities are nonzero, independence also gives P(A B)=P(A).

Conditioning restricts the sample space. For \(P(B)>0\), \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\), so \(P(A\cap B)=P(B)P(A\mid B)\). Divide by the probability of the given event, not the event being asked about.

Example: a bag contains \(3\) red and \(2\) blue balls. Without replacement, \(P(\text{two red})=\dfrac35\times\dfrac24=\dfrac3{10}\). With replacement and independent draws, it is \(\dfrac35\times\dfrac35=\dfrac9{25}\).

To test independence compare \(P(A\cap B)\) with \(P(A)P(B)\), or compare \(P(A\mid B)\) with \(P(A)\). If the relevant probabilities are nonzero, independence also gives \(P(A\mid B^{\prime})=P(A)\).

Probability

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Conditional Probability and Replacement

Conditional Probability and Replacement

Video Solution:

Solution:

Solution locked