Three coins are tossed once. Find the probability of: (a) 3 heads (b) exactly 2 heads (c) atleast 2 heads (d) atmost 2 heads (e) no tails (f) head and tail appear alternatively (g) atleast one head and one tail...
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Answers
Step-by-step explanation:
Three coins are tossed, so let S be the sample space: hence
S ={HHH, HHT, HTH, THH
TTT, TTH, THT, HTT}
n(S) = 8
(a) let A be the event of getting 3 heads:
A ={HHH}, n(A) = 1
P(A) = n(A)/n(S) =1/8
(b) let B be the event of getting exactly 2 heads:
B ={HHT, HTH, THH}, n(B) = 3
P(B) = n(B)/n(S) =3/8
(c) let C be the event of getting at least 2 heads:
C ={HHH, HHT, HTH, THH}, n(C) = 4
P(C) = n(C)/n(S) = 4/8 = 1/2
(d) let D be the event of getting atmost 2 heads:
D ={HHT, HTH, THH, TTT, TTH, THT, HTT}, n(D) = 7
P(D) = n(D)/n(S) = 7/8
(e) let E be the event of getting no tails:
E ={HHH}, n(E) = 1
P(E) = n(E)/n(S) =1/8
(f) let F be the event of head and tail appear alternatively
F ={HTH, THT}
n(F) = 2
P(F) = n(F)/n(S) = 2/8 = 1/4
(g) let G be the event of getting atleast one head and one tail
G = {HHT, HTH, THH, TTH, THT, HTT}
n(G) = 6
P(G) = n(G)/n(S) = 6/8 = 3/4
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