Total number of marbles = 10 + 20 + 30 = 60
Number of ways of drawing 5 marbles from 60 marbles =\(^{60}C_5\)
(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.
5 blue marbles can be drawn from 20 blue marbles in \(^{20}C_5 \)ways.
∴Probability that all marbles will be blue =\(\frac{^{20}C_5}{^{60}C_5}\)
(ii) Number of ways in which the drawn marble is not green =\(^{(20+10)}C_5=\space^{30}C_5\)
∴Probability that no marble is green =\( \frac{^{30}C_5}{^{60}C_5}\)
∴Probability that at least one marble is green =\(1- \frac{^{30}C_5}{^{60}C_5}.\)
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
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.