Three unbiased coins are tossed simultaneously. What is probability of getting at least two tails?
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Given: Number of unbiased coin = 3 Calculation: Total number of results when three coins are tossed = 23 = 8 Favourable result = (TTH, THT, HTT, TTT) Number of favourable results = 4 ∴ Probability of at least two heads = 4/8 = 1/2 Solution : In tossing three coins, the sample space is given gy <br> `S = {"HHH, HHT, HTH, THH, HTT, THT, TTH, TTT"}.` <br> And, therefore, n(S) = 8. <br> (i) Let `E_(1)` = event of getting all heads. Then, <br> `E_(1) = {HHH}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.` <br> (ii) Let `E_(2)` = event of getting 2 heads. Then, <br> `E_(2) = {HHT, HTH, THH}` and, therefore, `n(E_(2)) = 3.` <br> `therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.` <br> (iii) Let `E_(3)` = event of getting 1 head. Then, <br> `E_(3) = {"HTT, THT, TTH" }` and, therefore, `n(E_(3)) = 3.` <br> `therefore` P (getting 1 head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 3/8.` <br> (iv) Let `E_(4)` = event of getting at least 1 heads. Then, <br> `E_(4) = {"HTT, THT, TTH, HHT, HTH, THH, HHH"}` and, therefore, `n(E_(4)) = 7.` <br> `therefore` P (getting at least 1 head) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 7/8.` <br> (v) Let `E_(5)` = event of getting at least 2 heads. Then, <br> `E_(5) = {"HHT, HTH, THH, HHH"}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(5)) = (n(E_(5)))/(n(S)) = 4/8 = 1/2.` | Three unbiased coins are tossed together, then which of the following is true? A. The probability of getting exactly 2 heads is 12 B. The probability of getting atleast one head is 58 C. The probability of getting almost 2 tail is 38 D. The probability of getting exactly one tail is 38 Please scroll down to see the correct answer and solution guide. Right Answer is: DSOLUTIONSample Space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Related QuestionsLast updated date: 03rd Jan 2023 • Total views: 282.3k • Views today: 28.20k 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?Probability of getting no head = 3/8.
When 3 unbiased coins are tossed together once what is the probability of getting two heads?` <br> `therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.
When three unbiased coins are tossed simultaneously the number of possible outcomes are?Solution : When 3 coins are tossed simultaneously, all possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. <br> Total number of possible outcomes = 8.
When 3 unbiased coins are tossed once what is the probability of getting at least one head?Ben from St Peter's followed the tree diagram and calculated out the answer: If you flip a coin three times the chance of getting at least one head is 87.5%.
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