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.

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.

**Examples 1 :**

We tossed a single coin on the air and Find the probability of getting head ?

**Answer :**

Here n (S) is Total number of Possible outcomes So,

**S = { H, T }.**

**n(E) Total number of required outcomes. n(E)= 1{H}.**`

**P(E)Probability of Event.**

P(E) = n(E) / n(S) =** 1 / 2.**

**Example 2 :** Two Coins are impartial way throw on air and Find the Probability of 1 head

Answer :

Here n(S) is Total number of Possible outcomes So,

S = **{ H H, HT, TH, T T }.**

E = **Event of getting at most one head.**

E = **{T T,HT,TH}.**

P(E) = n(E) / n(S) =** 3 / 4.**

**Example 3:** We tossed simultaneously two coins. than find the probability of exactly one tail.

**Answer:** 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.

**Example 4:** We tossed three coins find the probability of no heads.

**Answer:** sample space S = { HHH, HHT, HTH, HTT, THT, TTH, THH, TTT }

Number of cases = 8

probability number of heads = probability number of Tails = 1 / 8.

**Example 5:** Three coins are tossed find the probability of at least two heads.

**Answer:** 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 = 4 / 8 = 1 / 2.

**Example 6:** We tossed four coins find the probability of four tails.

**Answer:** There are sixteen possibility of outcomes.

Sample space S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT,HTHT, HTTH, THTH,TTHH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}

Here is only favorable cases TTTT.

So, Probability of four tails is = 1 / 16.

**Example 7:**

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

**Answer :**

n ( S ) = 4 = { (T , T) , ( H , T ) , ( T , H ) , ( H , H ) }

probability of the appearing of at most one head = { HH , HT , TH , TT } ,

p ( E ) = 3 / 4 .

**Example 8:**

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

**Answer :**

n ( S ) = 4 = { (T , T), ( H , T ), ( T , H ), ( H , H ) }

probability of the appearing of at most two head = { HH , HT , TH , TT },

n ( E ) = 4 .

So , p ( E ) = 4 / 4 = 1.

- Probability problem on Balls
- Probability problem on Dice
- Probability Example 1
- Probability Example 2
- Probability Example 3
- Probability Example 4
- << Go back to Probability main page

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Dear Sir, In example 4 , how is answer 1

Ans 1/8