2 Fair coins are tossed simultaneously. So, the sample space(S) is given by :
S = {HH, HT, TH, TT}.
Let, A be the event that denotes HEAD on both the throws.
Let, B be the event that denotes HEAD on the first throw.
Let, C be the event that denotes HEAD on the second throw.
For, Event A (Head on both coins), only one case is feasible which is HH.
$\therefore$ P(A) = 1 / 4
For, Event B (Head on first coins), only two cases are feasible which is HH, HT.
$\therefore$ P(B) = 1 / 2
For, Event C (Head on second coins), only one case is feasible which is HH, TH.
$\therefore$ P(C) = 1 / 2
NOTE : Two or more events are said to be independent, when the occurrence of one event does not
affect the occurrence of other event.
In case of Independent Events : P(A $\cap$ B) = P(A) . P(B)
Let’s check options one by one :
Option – A) A & B are independent.
So, this will be true only when P(A $\cap$ B) = P(A) . P(B)
Here A $\cap$ B will have those cases having Head on first coin.
P(A $\cap$ B) = 1 / 2
P(A) = 1 / 4
P(B) = 1 / 2
P(A) . P(B) = 1 / 8
Clearly, P(A $\cap$ B) $\neq$ P(A) . P(B). So, Option A is FALSE.
Option – B) A & C are independent.
So, this will be true only when P(A $\cap$ C) = P(A) . P(C)
Here A $\cap$ C will have those cases having Head on second coin.
P(A $\cap$ C) = 1 / 2
P(A) = 1 / 4
P(C) = 1 / 2
P(A) . P(C) = 1 / 8
Clearly, P(A $\cap$ C) $\neq$ P(A) . P(C). So, Option B is FALSE.
Option – C) B & C are independent.
So, this will be true only when P(B $\cap$ C) = P(B) . P(C)
Here B $\cap$ C will have those cases having Head on both coin.
P(B $\cap$ C) = 1 / 4
P(B) = 1 / 2
P(C) = 1 / 2
P(B) . P(C) = 1 / 4
Clearly, P(B $\cap$ C) = P(A) . P(B). So, Option C is TRUE.
Option – D) P(B/C) = P(B)
P(B/C) = P(B $\cap$ C) / P(C)
P(B $\cap$ C) = 1 / 4 [See calculations in Option C]
P(C) = 1/2
P(B/C) = (1/4) / (1/2) = 2 / 4 = 1 / 2
P(B) = 1 / 2
Clearly, P(B/C) = P(B). So, Option D is TRUE.
Finally, we have correct Options : C & D.