a coin tossed three times . find the porbability of the following events (a) getting at least of two head (b) getting at most one head (c) getting more head than tails.
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Answer:
P(A)= 1 /2
P(B)=3/8
P(C)= 1/2
Step-by-step explanation:
Sample space: {HHH,HTH,THH,TTH
HHT,HTT,THT,TTT}
Total number of possible outcomes=8
(1) A getting at least two heads
P(A)=P(getting two heads)+P(getting 3 heads) = 3/8 + 1/8
(using formula=P(event)=
No. of favourable outcomes /Total no. of possible outcomes
P(A)= 4/8 = 1/2
∴P(A)= 1 /2
(2) B: getting exactly two heads
P(B)=3/8
(3) C: getting almost one head
P(C)=P(getting no head)+P(getting 1 head)
= 1/8 + 3/8 = 4/8 = 1/2
P(C)= 1/2
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