Qstnid -3- A Coin Is Tossed Twice. Find The Probability Of Getting: (I) Exactly One Head (Ii) Exactly One Tail (Iii) Two Tails (Iv) Two Heads - 23: Probability Class 8 Maths - Icse-viii-mathematics

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ICSE-VIII-Mathematics

23: Probability Class 8 Maths

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Qstn# 3 Prvs-Qstn Next-Qstn

#3
A coin is tossed twice. Find the probability of getting: (i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Ans : (i) Exactly one head
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(ii) Exactly one tail
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(iii) Two tails
Possible number of favourable outcomes = 1
(i.e., TT)
Total number of possible outcome s = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= ¼
(iv) Two heads
Possible number of favourable outcomes = 1
(i.e. HH)
Total number of possible outcomes = 4
∴ P(E) = ¼

#3-i
exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Ans : Exactly one head
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(ii) Exactly one tail
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(iii) Two tails
Possible number of favourable outcomes = 1
(i.e., TT)
Total number of possible outcome s = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= ¼
(iv) Two heads
Possible number of favourable outcomes = 1
(i.e. HH)
Total number of possible outcomes = 4
∴ P(E) = ¼