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
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Answer
Hint- In order to solve such type of question we must use formula Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] , along with proper understanding of favourable cases and total cases.Complete step-by-step answer:If two unbiased coins are tossed simultaneously, then the total number of possible outcomes may be either.1.both head HH2.one head and one tail (HT, TH)3.Both tail (TT)So here total number of possible cases = 4(i) two headsa favourable outcome for two heads is HH.No. of favourable outcomes = 1total number of possible cases = 4As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] Probability (two heads) =$\dfrac{1}{4}$(ii) favourable outcomes for one tail are TH, HT.No. of favourable outcomes = 2total number of possible cases = 4As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] Probability(one tail) = $\dfrac{2}{4}{\text{ = }}\dfrac{1}{2}$(iii) favourable outcomes for at least one tail are TH, HT, TT.No. of favourable outcomes = 3total number of possible cases = 4As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] Probability (at least one tail) = $\dfrac{3}{4}$ (iv) Favourable outcomes for at least one head are TH, HT, HH.No. of favourable outcomes = 3total number of possible cases = 4As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] Probability (at least one head) = $\dfrac{3}{4}$(v) favourable outcomes for no tail means not a single head is HH.No. of favourable outcomes = 1total number of possible cases = 4As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\] Probability(no tail) = $\dfrac{1}{4}$
NOTE- In Such Types of Question first find out the total numbers of possible outcomes then find out the number of favourable cases, then divide them using the formula which stated above, we will get the required answer.