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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10. Two coins are tossed together. What is the probability of getting different faces on the coins?
According to the question it is given that, two coins are tossed together. As we all know that, coin has two faces one is head and other one is tail. The possibilities of getting head and tail when two coins are tossed = {HH, TT, HT, TH} Total number of outcomes $=4$ Number of favorable outcomes $=2$ P(getting different faces on the coins) = Number of favorable outcomes/total number of Outcomes $=2/4$ $=1/2$ Two coins are tossed together . what is the probabilty of getting different faces on the coins ? The sample space for the experiment of tossing two coins is {HH , HT , TH , TT} Number of outcomes = 4Number of favourable outcomes = 2 P(getting different faces on the coins) = `"Number of favourable outcomes"/"Total number of outcomes"` = `2/4 = 1/2` Concept: Basic Ideas of Probability Is there an error in this question or solution? |