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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The probability that the first and third coins are heads is $1/3\times 1/4=1/12$. The probability that the first coin is heads is $2/3\times 1/2=1/3$. So the conditional probability that the third coin is heads, given that the first coin is heads, is $(1/12) \div(1/3)=1/4$. As for the second part, the total number of heads is $0$ with probability $1/4$, $1$ with probability $1/2$, and $2$ with probability $1/4$. From this, you can find $$ E[X]=\frac{1}{4}\cdot 0+\frac{1}{2}\cdot 1 + \frac{1}{4}\cdot 2=1; $$ and $$ E[X^2]=\frac{1}{4}\cdot 0^2+\frac{1}{2}\cdot 1^2 + \frac{1}{4}\cdot 2^2=\frac{3}{2}, $$ so $\text{Var}[X]=E[X^2]-E[X]^2=3/2-1=1/2.$ Which of the following is an equation representing a redox reaction? Sana maging okay na yung lahat kasi nakakadepress na talaga. Chos, tamad lang pala ako mag modules:') Which of the following graphs shows that the object’s motion is accelerating? Sino jn pede maging couple ko? Sa pic lng po ah Answer the following questions base on the given figure. write your answer on space provided. In the adjoining figure, what term must be used to describe each of the following: for 31-35, refer to the figure below:31. using the formula for the perimeter, solve for x.A. 10B. 15C. 20D. 2532.What does the x in the figure represe … Can someone help me finish my homework?:( 13. A shoe box measures 30 cm long, 18 cm wide and 12 cm deep. What is the amount of space inside the box? Which of the following trigonometry ratios of ∅ in the given figure below has the value of 2/3? |