Elementary events associated to random experiment of tossing three coins are Show
HHH , HHT, HTH ,THH , HTT,THT,TTH ,TTT Total number of elementary events = 8. (i) The event getting all heads is said to occur, if the elementary event HHH occurs, i.e.HHH is an outcome. Favourable number of elementary events = 1 Hence, required probability =1/8 (ii) The event "getting two heads" will occur, if one of the elementary events HHT, THH, HTH occurs. Favourable number of elementary events =3 Hence, required probability=3/8 (iii) The event of "getting one head", when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs. Favourable number of elementary events = 3 Hence, required probability=3/8 (iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event "getting at least two heads" occurs. Favourable number of elementary events = 4 Hence, required probability=4/8=1/2. (v) Similar as (iv) P (getting at least two tails)=4/8=1/2. Open in App Solution When 3 unbaised coins are tossed, the total number of events = 8 They are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} (1) Number of favourable events (getting at least two heads) = 4 Probability of getting at least two heads = 4/8 = 1/2 (adsbygoogle = window.adsbygoogle || []).push({}); (2) At most two heads That is no head or one head or 2 heads Probability of getting at most two heads = 7/8 3) Two head = 3 (HHT,HTH,THH) Probability of getting no head = 3/8.Answer Verified
Hint: To solve this problem we need to first write the sample space of all the possible outcomes and then check the favorable outcomes that satisfy the given condition. After getting the favorable outcomes and total number of possible outcomes substitute them in the probability formula. Complete step-by-step answer: Note: While writing the sample space one needs to make sure that you consider all the possible outcomes without fail because neglecting any one of the possible outcomes causes the entire result to be wrong. When 3 unbiased coins are tossed once what is the probability of getting all heads?` <br> `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.
When 3 unbiased coins are tossed once what is the probability of getting at least one tail?Solution: When 3 coins are tossed, the possible outcomes are HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. (i) Let E1 denotes the event of getting all tails. Hence the required probability is ⅛.
When 3 unbiased coins are tossed once what is the probability of getting at least 2 heads?∴ Probability of getting at least two heads are 1/2.
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