Let me work through your reasoning and demonstrate exactly why it doesn't work. Just now, I asked my friend to flip 3 coins, one at a time. I then asked her if at least one of the coins came up tails. She said yes. What's the probability that all three coins came up tails? Define $A$ as the first coin which came up tails. Define $B$ as the first coin which is not $A$, and then define $C$ has the remaining coin. If I make your reasoning more explicit, it says something like:
However, this reasoning isn't correct. We have defined $B$ in an "unfair" way, so that $B$ is actually more likely to be heads than tails. (What's so unfair about the definition of $B$? If the first coin comes up heads, then $B$ comes up heads, because, according to our definition, $B$ is the first coin. But if the first coin comes up tails, $B$ may come up heads anyway.) We can see the actual probabilities of $B = T$ and $C = T$ by writing out the seven possible outcomes (all of which are equally likely):
By looking at this table, we can see that $P(B = T) = 2/7$ and $P(C = T) = 3/7$. The two events aren't independent, either: $P(B = T)P(C = T) = 6/49$, whereas $P(B = T, C = T) = 1/7 = 7/49$.
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 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.
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