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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. The 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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The ratio of successful events A = 120 to the total number of possible combinations of a sample space S = 128 is the probability of 2 heads in 7 coin tosses. Users may refer the below solved example work with steps to learn how to find what is the probability of getting at-least 2 heads, if a coin is tossed seven times or 7 coins tossed together. Users may refer this tree diagram to learn how to find all the possible combinations of sample space for flipping a coin one, two, three or four times. Solution Step by step workout step 2 Find the expected or successful events A A = {HHHHHHH, HHHHHHT, HHHHHTH, HHHHHTT, HHHHTHH, HHHHTHT, HHHHTTH, HHHHTTT, HHHTHHH, HHHTHHT, HHHTHTH, HHHTHTT, HHHTTHH, HHHTTHT, HHHTTTH, HHHTTTT, HHTHHHH, HHTHHHT, HHTHHTH, HHTHHTT, HHTHTHH, HHTHTHT, HHTHTTH, HHTHTTT, HHTTHHH, HHTTHHT, HHTTHTH, HHTTHTT, HHTTTHH, HHTTTHT, HHTTTTH, HHTTTTT, HTHHHHH, HTHHHHT, HTHHHTH, HTHHHTT, HTHHTHH, HTHHTHT, HTHHTTH, HTHHTTT, HTHTHHH, HTHTHHT, HTHTHTH, HTHTHTT, HTHTTHH, HTHTTHT, HTHTTTH, HTHTTTT, HTTHHHH, HTTHHHT, HTTHHTH, HTTHHTT, HTTHTHH, HTTHTHT, HTTHTTH, HTTHTTT, HTTTHHH, HTTTHHT, HTTTHTH, HTTTHTT, HTTTTHH, HTTTTHT, HTTTTTH, THHHHHH, THHHHHT, THHHHTH, THHHHTT, THHHTHH, THHHTHT, THHHTTH, THHHTTT, THHTHHH, THHTHHT, THHTHTH, THHTHTT, THHTTHH, THHTTHT, THHTTTH, THHTTTT, THTHHHH, THTHHHT, THTHHTH, THTHHTT, THTHTHH, THTHTHT, THTHTTH, THTHTTT, THTTHHH, THTTHHT, THTTHTH, THTTHTT, THTTTHH, THTTTHT, THTTTTH, TTHHHHH, TTHHHHT, TTHHHTH, TTHHHTT, TTHHTHH, TTHHTHT, TTHHTTH, TTHHTTT, TTHTHHH, TTHTHHT, TTHTHTH, TTHTHTT, TTHTTHH, TTHTTHT, TTHTTTH, TTTHHHH, TTTHHHT, TTTHHTH, TTTHHTT, TTTHTHH, TTTHTHT, TTTHTTH, TTTTHHH, TTTTHHT, TTTTHTH, TTTTTHH} A = 120step 3 Find the probability P(A) = Successful Events/Total Events of Sample Space = 120/128 = 0.94 P(A) = 0.94 0.94 is the probability of getting 2 Heads in 7 tosses. |
7 coins are tossed simultaneously what is the probability of getting at least two heads
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