Determine P: A coin is tossed three times, where
A coin tossed three times, i.e., S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)
⇒ n(S)=8
(i) E: heads on third toss E=(TTH, HTH, THH, HHH)
⇒N(E)=4
\(P(E)=\frac {n(E)}{n(S)}\) = \(\frac 48\) =\(\frac 12\)
F: heads on first two toss F=(HHT, HHH)
⇒ n(F)=2
\(P(F)=\frac {n(F)}{n(S)}\) =\(\frac 28\) =\(\frac 14\)
∴E∩F = (HHH)
⇒ n(E∩F) = 1
∴\(P(E∩F)=\frac {n(E∩F)}{n(S)} =\frac 18\)
And, \(P(E|F)=\frac {P(E∩F)}{P(F)} =\frac {1/8}{2/8} =\frac 12\)
(ii )E: at least two heads E=(HHT, HTH, THH, HHH)
⇒n(E)=4
\(P(E)=\frac {n(E)}{n(S)}\) =\(\frac 48\) =\(\frac 12\)
F: at most two heads F=(TTT, HTT, THT, TTH, HHT, HTH, THH)
⇒n(F)=7
\(P(F)=\frac {n(F)}{n(S)}\) =\(\frac 78\)
∴E∩F = (HHT, HTH, THH)
⇒n(E∩F) = 3
\(∴P(E∩F) = \frac {n(E∩F)}{n(S)}\) =\(\frac 38\)
And, \(P(E|F)=\frac {P(E∩F)}{P(F)} =\frac {3/8}{7/8} =\frac 37\)
(iii)E: at most two tails E=(HTT,THT,TTH,HHT,HTH,THH,HHH)
⇒n(E)=7
\(P(E)=\frac {n(E)}{n(S)}\) =\(\frac 78\)
F:at least one tail F = (TTT, HTT, THT, TTH, HHT, HTH, THH)
⇒n(F)=7
\(P(F)=\frac {n(F)}{n(S)}\) =\(\frac 78\)
∴E∩F=(HTT, THT, TTH, HHT, HTH, THH)
⇒n(E∩F)=6
∴\(P(E∩F)=\frac {n(E∩F)}{n(S)}\) =\(\frac 68\)
And, \(P(E∩F)=\frac {n(E∩F)}{n(S)}\)=\(\frac {6/8}{7/8}\) =\(\frac 67\)
Rupal, Shanu and Trisha were partners in a firm sharing profits and losses in the ratio of 4:3:1. Their Balance Sheet as at 31st March, 2024 was as follows:
(i) Trisha's share of profit was entirely taken by Shanu.
(ii) Fixed assets were found to be undervalued by Rs 2,40,000.
(iii) Stock was revalued at Rs 2,00,000.
(iv) Goodwill of the firm was valued at Rs 8,00,000 on Trisha's retirement.
(v) The total capital of the new firm was fixed at Rs 16,00,000 which was adjusted according to the new profit sharing ratio of the partners. For this necessary cash was paid off or brought in by the partners as the case may be.
Prepare Revaluation Account and Partners' Capital Accounts.
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.