You're trying to find the right way to apply the principle of indifference. It's not entirely unreasonable to consider applying it to the number of tails thrown. In a Bayesian context, if you didn't know how to calculate the probability distribution for the number of tails, you might assign a subjective probability of $\frac14$ to each possibility since you have no way of judging one to be more probable than another. However, this would only be a reflection of your subjective beliefs. There's no objective symmetry in the situation that dictates that the different numbers of tails should be equiprobable. By contrast, there's a manifest symmetry between heads and tails (disregarding small differences between the designs on the two sides of the coins), which dictates that these two results must be objectively equiprobable. This is part of the implicit premise of the question that these are fair coins. If it weren't part of what you know about the situation that the coins are fair, and you had no way of knowing to which side the coins might be biased, you could still apply the principle of indifference and assign the same probability to the two outcomes – but that would then again be an assignment of subjective probabilities based on your subjective beliefs. Since in the present case we do assume that you know that the coins are fair, the objective symmetry between heads and tails trumps whatever subjective inclination you may have to consider the numbers of tails equiprobable. As you've shown using the $8$ different combinations, the different numbers of tails are not equiprobable, and once you know this, there's no longer any reason to apply the principle of indifference to them – you're no longer indifferent, and there's no symmetry between them that would force you to be indifferent and would produce a contradiction between the two different applications of the principle of indifference.
Here we will learn how to find the probability of tossing two coins. Let us take the experiment of tossing two coins simultaneously: When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail. Therefore, total numbers of outcome are 22 = 4The above explanation will help us to solve the problems on finding the probability of tossing two coins. Worked-out problems on probability involving tossing or flipping two coins: 1. Two different coins are tossed randomly. Find the probability of: (i) getting two heads (ii) getting two tails (iii) getting one tail (iv) getting no head (v) getting no tail (vi) getting at least 1 head (vii) getting at least 1 tail (viii) getting atmost 1 tail (ix) getting 1 head and 1 tail Solution: When two different coins are tossed randomly, the sample space is given by S = {HH, HT, TH, TT} Therefore, n(S) = 4. (i) getting two heads: Let E1 = event of getting 2 heads. Then,E1 = {HH} and, therefore, n(E1) = 1. Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4. (ii) getting two tails: Let E2 = event of getting 2 tails. Then,E2 = {TT} and, therefore, n(E2) = 1. Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4. (iii) getting one tail: Let E3 = event of getting 1 tail. Then,E3 = {TH, HT} and, therefore, n(E3) = 2. Therefore, P(getting 1 tail) = P(E3) = n(E3)/n(S) = 2/4 = 1/2 (iv) getting no head: Let E4 = event of getting no head. Then,E4 = {TT} and, therefore, n(E4) = 1. Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = ¼. (v) getting no tail: Let E5 = event of getting no tail. Then,E5 = {HH} and, therefore, n(E5) = 1. Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼. (vi) getting at least 1 head: Let E6 = event of getting at least 1 head. Then,E6 = {HT, TH, HH} and, therefore, n(E6) = 3. Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾. (vii) getting at least 1 tail: Let E7 = event of getting at least 1 tail. Then,E7 = {TH, HT, TT} and, therefore, n(E7) = 3. Therefore, P(getting at least 1 tail) = P(E2) = n(E2)/n(S) = ¾. (viii) getting atmost 1 tail: Let E8 = event of getting atmost 1 tail. Then,E8 = {TH, HT, HH} and, therefore, n(E8) = 3. Therefore, P(getting atmost 1 tail) = P(E8) = n(E8)/n(S) = ¾. (ix) getting 1 head and 1 tail: Let E9 = event of getting 1 head and 1 tail. Then,E9 = {HT, TH } and, therefore, n(E9) = 2. Therefore, P(getting 1 head and 1 tail) = P(E9) = n(E9)/n(S)= 2/4 = 1/2. The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins. Probability Probability Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability of Tossing Two Coins to HOME PAGE
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Here we will learn how to find the probability of tossing three coins. Let us take the experiment of tossing three coins simultaneously: When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail. Therefore, total numbers of outcome are 23 = 8The above explanation will help us to solve the problems on finding the probability of tossing three coins. Worked-out problems on probability involving tossing or throwing or flipping three coins: 1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times. If three coins are tossed simultaneously at random, find the probability of: (i) getting three heads, (ii) getting two heads, (iii) getting one head, (iv) getting no head Solution: Total number of trials = 250. Number of times three heads appeared = 70. Number of times two heads appeared = 55. Number of times one head appeared = 75. Number of times no head appeared = 50. In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,(i) getting three heads P(getting three heads) = P(E1) Number of times three heads appeared= Total number of trials = 70/250 = 0.28 (ii) getting two heads P(getting two heads) = P(E2) Number of times two heads appeared= Total number of trials = 55/250 = 0.22 (iii) getting one head P(getting one head) = P(E3) Number of times one head appeared= Total number of trials = 75/250 = 0.30 (iv) getting no head P(getting no head) = P(E4) Number of times on head appeared= Total number of trials = 50/250 = 0.20 Note: In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4)= (0.28 + 0.22 + 0.30 + 0.20) = 1 2. When 3 unbiased coins are tossed once. What is the probability of: (i) getting all heads (ii) getting two heads (iii) getting one head (iv) getting at least 1 head (v) getting at least 2 heads (vi) getting atmost 2 heads Solution: In tossing three coins, the sample space is given by S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} And, therefore, n(S) = 8. (i) getting all heads Let E1 = event of getting all heads. Then,E1 = {HHH} and, therefore, n(E1) = 1. Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8. (ii) getting two heads Let E2 = event of getting 2 heads. Then,E2 = {HHT, HTH, THH} and, therefore, n(E2) = 3. Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8. (iii) getting one head Let E3 = event of getting 1 head. Then,E3 = {HTT, THT, TTH} and, therefore, n(E3) = 3. Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8. (iv) getting at least 1 head Let E4 = event of getting at least 1 head. Then,E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH} and, therefore, n(E4) = 7. Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8. (v) getting at least 2 heads Let E5 = event of getting at least 2 heads. Then,E5 = {HHT, HTH, THH, HHH} and, therefore, n(E5) = 4. Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2. (vi) getting atmost 2 heads Let E6 = event of getting atmost 2 heads. Then,E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT} and, therefore, n(E6) = 7. Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8 3. Three coins are tossed simultaneously 250 times and the outcomes are recorded as given below.
If the three coins are again tossed simultaneously at random, find the probability of getting (i) 1 head (ii) 2 heads and 1 tail (iii) All tails Solution: (i) Total number of trials = 250. Number of times 1 head appears = 100. Therefore, the probability of getting 1 head = \(\frac{\textrm{Frequency of Favourable Trials}}{\textrm{Total Number of Trials}}\) = \(\frac{\textrm{Number of Times 1 Head Appears}}{\textrm{Total Number of Trials}}\) = \(\frac{100}{250}\) = \(\frac{2}{5}\) (ii) Total number of trials = 250. Number of times 2 heads and 1 tail appears = 64. [Since, three coins are tossed. So, when there are 2 heads, there will be 1 tail also]. Therefore, the probability of getting 2 heads and 1 tail = \(\frac{\textrm{Number of Times 2 Heads and 1 Trial appears}}{\textrm{Total Number of Trials}}\) = \(\frac{64}{250}\) = \(\frac{32}{125}\) (iii) Total number of trials = 250. Number of times all tails appear, that is, no head appears = 38. Therefore, the probability of getting all tails = \(\frac{\textrm{Number of Times No Head Appears}}{\textrm{Total Number of Trials}}\) = \(\frac{38}{250}\) = \(\frac{19}{125}\). These examples will help us to solve different types of problems based on probability of tossing three coins.
Probability Probability Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability of Tossing Three Coins to HOME PAGE
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