Thủ Thuật Hướng dẫn Two coins are tossed simultaneously the probability of getting most one head is Chi Tiết
Bùi Trung Huấn đang tìm kiếm từ khóa Two coins are tossed simultaneously the probability of getting most one head is được Cập Nhật vào lúc : 2022-11-23 08:14:06 . Với phương châm chia sẻ Bí quyết Hướng dẫn trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tham khảo Post vẫn ko hiểu thì hoàn toàn có thể lại Comment ở cuối bài để Mình lý giải và hướng dẫn lại nha.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 = 4
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 least 1 head
(vii) getting 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 = sự kiện 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 = sự kiện 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 = sự kiện 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 = sự kiện 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 = sự kiện 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 least 1 head:
Let E6 = sự kiện of getting least 1 head. Then,
E6 = HT, TH, HH and, therefore, n(E6) = 3.
Therefore, P(getting least 1 head) = P(E6) = n(E6)/n(S) = ¾.
(vii) getting least 1 tail:
Let E7 = sự kiện of getting least 1 tail. Then,
E7 = TH, HT, TT and, therefore, n(E7) = 3.
Therefore, P(getting least 1 tail) = P(E2) = n(E2)/n(S) = ¾.
(viii) getting atmost 1 tail:
Let E8 = sự kiện 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 = sự kiện 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
Verified
Hint: First, find the sample space of the problem. Next, find the number of elements in the sample space = total number of outcomes. Now, let E be an sự kiện when the head appears on one coin and the tail on another coin. Find the set E and find the number of elements in the sample space = total number of outcomes. Next, use the formula: $P(E)=dfractextnumber of favourable outcomestexttotal number of outcomes$ to get the final answer.
Complete step-by-step answer:
In this question, we are given that two fair coins are tossed simultaneously.
We need to find the probability of getting Head on one coin and Tail on the other coin.
When two coins are tossed simultaneously, the sample space is given by : S = HH, HT, TH, TT where, H is the appearance of Head and T is the appearance of the Tail on the coin.
So, the number of elements in the sample space = total number of outcomes is given by the following:
n (S) = 4
Let E be an sự kiện when
the head appears on one coin and the tail on another coin. E:HT,TH
So, the number of elements in the E = number of favourable outcomes is given by the following:
n (E) = 2
Now, we know that probability of an sự kiện is equal to the number of favourable outcomes divided by the total number of outcomes. i.e. $P(E)=dfractextnumber of favourable outcomestexttotal number of outcomes$
Using the above formula, we will get the following:
$P(E)=dfrac24=dfrac12$
Hence, the probability of getting Head on one coin and Tail on the other coin is equal to $dfrac12$.
This is the final answer.
Note: In this question, it is very important to know what a sample space is. In probability theory, the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment.
What is the probability of most one head?
Thus, most one head means number of heads in probability favourable outcomes is 0 or 1 .What is the probability of getting least one head when a coin is tossed twice?
If the probability of getting no heads is 14 , the probability of getting least one head is 1−41=43. Tải thêm tài liệu liên quan đến nội dung bài viết Two coins are tossed simultaneously the probability of getting most one head is