I have now thought about this a bit more, and I think the way we get the information "one gets head" is important.
In the question, I put that the possible cases were:
- Head - Head
- Head - Tail
- Tail - Head
- Tail - Tail
Each tends to happen $25$% of the time.
If we see the first to fall comes up head, then we restrict to cases $1$ and $2$, and so the probabilities are $1/2$. Only in $50$% of the time we can see the first one coming up head, and in its half both coins end having head.
If the information was gotten from another person who sees the outcomes while you have not, and that person is forced to tell you that one coin came up head always that it happens, then that person will do it $75$% of the time, of which its $1/3$ is when both coins get head. So, once we know we are in those $75$%, we have $1/3$ probability.
If that another person was free to say you either "one coin came up head" or "one coin came up tail", then the probabilities are $1/2$ for both coins coming up head. The reason is that in the two cases we have got a head and a tail, that person has $1/2$ probability to report "one coin came up head".
$1)$ $25$% of the time we get two heads, and in all this cases that person tells you that one coin came up head.
$2)$ $50$% of the time we get a head and a tail, but only in half of its cases ($25$%), he tells you that one of them came up head. In the other half, he tells you that one of them came up tail.
$3)$ $25$% of the time we get two tails, and in all this cases that person tells you that one coin came up tail.
So, only in $25$% $+ 25$% $= 50$% he says you that one coin came up head, of which its half is when you have both coins head.
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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.
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:
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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