Probability problem on Coin shortcut tricks are very important thing to know for your exams. Competitive exams are all about time. If you know how to manage time then you will surely do great in your exam. Most of us miss this thing. We provide examples on Probability problem on Coin shortcut tricks here in this page below. All tricks on probability problem on coin are provided here. We request all visitors to read all examples carefully. These examples will help you to understand shortcut tricks on Probability problem on Coin.

Before starting anything just do a math practice set. Write down twenty math problems related to this topic on a page. Do first ten maths using basic formula of this math topic. You also need to keep track of timing. After solving all ten math questions write down total time taken by you to solve those questions. Now practice our shortcut tricks on probability problem on coin and read examples carefully. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of Timing. You will surely see the improvement in your timing this time. But this is not all you want. You need more practice to improve your timing more.

Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. That doesn’t mean that other topics are less important. But only math portion can leads you to a good score. You can get good score only by practicing more and more. The only thing you need to do is to do your math problems correctly and within time, and this can be achieved only by using shortcut tricks. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may have that potential that you may do maths within time without using any shortcut tricks.

But, other peoples may not do the same. Here we prepared probability problem on coin shortcut tricks for those people. We try our level best to put together all types of shortcut methods here. But it possible we miss any. We appreciate if you share that with us. Your help will help others.

Here is some Probability on coin Examples are given, Before going through this examples u should remember all probability formula and fact that are required here for solved the Example, Let do the Problems on Probability on coin.

Example #1 – Probability problem on Coin

If we toss a single coin on the air, what will be the probability of getting head?

0/2

1/2

2/2

1/1

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve Here n(S) is total number of possible outcomes. So, S = { H, T }.

n(E) is total number of required outcomes. So, n(E) = 1{H}.

And, P(E) is probability of event. P(E) = n(E) / n(S) = 1/2.

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Example #2 – Probability problem on Coin

Two coins are impartial way throw on air and find the probability of at least one head.

1/4

2/4

3/4

4/4

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Answer: Option (C)

How to Solve Here n(S) is total number of possible outcomes. So, S = { HH, HT, TH, TT }.

E = Event of getting at most one head. E = { HH, HT, TH }.

So, P(E) = n(E) / n(S) = 3/4.

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Example #3 – Probability problem on Coin

If we tossed simultaneously two coins. Find the probability of exactly one tail.

1/2

2/2

2/3

3/4

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Answer: Option (A)

How to Solve Sample space S = { HH, HT, TH, TT } Number of cases = 4. Favorable cases HT, TH. So, probability of exactly one tail, = 2/4 = 1/2.

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Example #4 – Probability problem on Coin

If we tossed three coins simultaneously, then find the probability of no heads.

1/4

1/8

2/4

2/8

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve Sample space S = { HHH, HHT, HTH, HTT, THT, TTH, THH, TTT } Number of cases = 8. Probability number of no heads = probability number of tails = 1/8.

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Example #5 – Probability problem on Coin

Three coins are tossed find the probability of at least two heads.

1/2

1/4

1/8

2/8

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve sample space S = { HHH, HHT, HTH, HTT, THT, TTH, THH, TTT } Number of cases = 8 Favorable cases = HHT, HTH, THH, HHH. probability of at least two head is, = 4/8 = 1/2.

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Example #6

We tossed four coins find the probability of four tails.

1/2

1/4

1/8

1/16

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Answer: Option (D)

How to Solve There are sixteen possibility of outcomes. Sample space S = { HHHH, HHHT, HHTH, HTHH, THHH, HHTT,HTHT, HTTH, THTH, TTHH, THHT, TTTH, TTHT, THTT, HTTT, TTTT } Here is only favorable cases TTTT. So, probability of four tails is, = 1/16.

Rough Workspace

Example #7

If two coins are tossed, then what is the probability of the appearing of at most one head?

1/4

2/4

3/4

4/4

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (C)

How to Solve n( S ) = 4 = { TT, HT, TH, HH } Probability of the appearing of at most one head is, = { TT, HT, TH } p( E ) = 3/4.

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Example #8

Two coins are tossed. What is the probability of appearing at most two head?

1/4

2/4

3/4

4/4

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (D)

How to Solve n( S ) = 4 = { TT, HT, TH, HH } Probability of the appearing of at most two head is, = { TT, HT, TH, HH }, n( E ) = 4. So, p( E ) = 4/4.

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Sample Space {(H,H), (H,T), (T,H), (T,T)} P(E) = n(E)/n(S) P(E) = !/4 where, P(E) = Probability of event n(E) = No of required outcome n(S) = No of possible outcomes

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Dear Sir, In example 4 , how is answer 1

Ans 1/8

Sample Space {(H,H), (H,T), (T,H), (T,T)}

P(E) = n(E)/n(S)

P(E) = !/4

where,

P(E) = Probability of event

n(E) = No of required outcome

n(S) = No of possible outcomes

4th question asking no head, no head means only tails (TTT) so 3 coins total 2power3 = 8 only tails are 1 possibility so finally 1/8

Eslia kyuki usne bola ek bhi head na hone ki probability batani ha to wo to sirf tabhi hogi jab TTT aayega

Can you tell me how to solve 8 coins problem its not possible to count hh,tt all d time

its is possible

2^8=256, so there are 256 possibilities. Say you want exactly three heads. The probability would be (8C3)((1/2)^3)((1/2)^5)

Sir how to solve at most coin tossing like 7 th and 8 th sum