Math, asked by susanna1, 1 year ago

3 coins are tossed together . find the probability of getting : 2 heads. at least 2 heads. at most 2 heads. at most 1 tail. at least 1 tail. 3 heads.



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Answers

Answered by siddhartharao77
69

Total number of coins tossed = 3.

Total number of outcomes n(S) = 8.

(i) Let A be the event of getting 2 heads:

= > n(A) = {HTH,HHT,THH} = 1.

Required probability p(A) = n(A)/n(S) = 3/8.

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(ii) Let B be the event of getting atleast 2 heads.

= > n(B) = {HHH,HHT,HTH,THH}.

Required probability P(A) = n(B)/n(S) = 4/8 = 1/2.

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(iii) Let C be the event of getting at most 2 heads.

= > n(C) = {HHT,HTH,THH,TTH,THT,HTT,TTT} = 7

Required probability P(C) = 7/8

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(iv) Let D be the event of getting at most 1 tail.

= > n(D) = {HHH,THH,HTH,HHT} = 4.

Required probability P(D) = n(D)/n(S) = 4/8.

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(v) Let E be the event of getting atleast one tail:

= > n(E) = {HHT,HTH,HTT,TTT,THT,TTH,THH} = 7.

Required probability P(E) = n(E)/n(S) = 7/8

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(vi) Let F be the event of Getting 3 heads.

= > n(F) = {HHH}.

Required probability P(F) = n(F)/n(S) = 1/8.

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Hope it helps!

Answered by sneha9143
8

Answer:

3/8

Step-by-step explanation:

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