Conditional probability class 12th ncert maths latest syllabus 2026-2027 | By Ashish Sir

Описание: Conditional probability class 12th ncert maths latest syllabus 2026-2027 | By Ashish Sir
‪@Integralganit‬

Your Queries:-
ln the video the number question completed i.e given below:-

6. A coin is tossed three times, where
(i) E : head on third toss , F : heads on first two tosses
(ii) E : at least two heads , F : at most two heads
(iii) E : at most two tails , F : at least one tail,

7. Two coins are tossed once, where
(i) E : tail appears on one coin, F : one coin shows head
(ii) E : no tail appears, F : no head appears,

8. A die is thrown three times,
E : 4 appears on the third toss, F : 6 and 5 appears respectively
on first two tosses,

9. Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle,

10. A black and a red dice are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given
that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4.,

11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5}
Find
(i) P(E|F) and P(F|E) (ii) P(E|G) and P(G|E)
(iii) P((E ∪ F)|G) and P((E ∩ F)|G),

The salient features of the chapter are –
The conditional probability of an event E, given the occurrence of the event F
is given by P(E F) P(E | F)
P(F)
∩
= , P(F) ≠ 0,

0 ≤ P (E|F) ≤ 1, P (E′|F) = 1 – P (E|F)
P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G),

P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0
P (E ∩ F) = P (F) P (E|F), P (F) ≠ 0,

If E and F are independent, then
P (E ∩ F) = P (E) P (F)
P (E|F) = P (E), P (F) ≠ 0
P (F|E) = P (F), P(E) ≠ 0,

Theorem of total probability
Let {E1
, E2
, ...,En
) be a partition of a sample space and suppose that each of
E1
, E2
, ..., En
has nonzero probability. Let A be any event associated with S,
then
P(A) = P(E1
) P (A|E1
) + P (E2
) P (A|E2
) + ... + P (En
) P(A|En
),

Bayes' theorem If E1
, E2
, ..., En
are events which constitute a partition of
sample space S, i.e. E1
, E2
, ..., En
are pairwise disjoint and E1 4 E2 4 ... 4 En
= S
and A be any event with nonzero probability, then
P(E A P(E ) P(A|E )
P(E )P(A|E )
i i
i
j j
j
n
| )=
=
∑
1
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