**Probability**To understand probability more accurately, let us understand an case of rolling a dice, the potential outcomes are – 1, 2, 3, 4, 5, and 6. The probability of happening any of the probably affairs is 1/6. As the possibility of happening any of the affairs is the lapp so there is an equal possibility of happening any favorable affair, in this casing, it is either of two 1/6 or 50/3. formula of Probability

P ( A ) = { Number of favorable affair to A } ⁄ { full number of affair }

**Terms related to probability**

**Experiment:**Any functioning that gives a well-defined result is known as an experiment. For example: Flipping a coin or tossing a die is an experiment.**Random Experiment:**In any experiments, all likely affair but one does not know which exact affair will happen. This is called a Random experiment. For example: By flipping a coin, either heads or tails are acquired but one is not sure that only the head will occur or the tail will occur.**Sample Space:**Sample space is the group of all likely events. Example: On flipping a coin we have 2 results: heads and tails.**Trial:**It is a process by which the experiment is executed and the result is acclaimed. For example: choosing a card from a deck of 52 cards.**Event:**Each result of an experiment is called an affair or event. For example: Getting a tail on flipping a coin is an affair or event.**Independent Events:**When the happening of one affair is not affected by the happening of another affair then it is known as independent events. For Example, one can concurrently flip a coin and throw a dice as they are unconnected affairs.**Exhaustive Events:**Two events or affairs are said to be exhaustive if their joining is equal to sample space.**Exclusive Events:**When two affairs cannot happen at the same time or the two affairs are disjoint, they are said to be exclusive events. For Example: On flipping, a coin one can get either head or tail but not both.

### What is the probability sample space of tossing 4 coins?

**Solution:**

Each coin flip has 2 probable events, so the flip of 4 coins has 2×2×2×2 = 16 likely events. We can summarize all probable events as follows, where H shows a headway, and T a tail : HHHH THHH HHHT THHT HHTH THTH HHTT THTT HTHH TTHH HTHT TTHT HTTH TTTH HTTT TTTT If we suppose that each single coin is equally probable to come up heads or tails, then each of the above 16 result to 4 flips is equally probable. Each happens a fraction one out of 16 times, or each has a possibility of 1/16. On the other hand, we could assert that the 1st coin has hypothesis 1/2 to come up heads or tails, the 2nd coin has possibility 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the hypothesis for any one particular order of heads and tails is merely ( 1/2 ) × ( 1/2 ) × ( 1/2 ) × ( 1/2 ) = ( 1/16 ). now lets necessitate : what is the hypothesis that in 4 flips, one gets N heads, where N = 0, 1, 2, 3, or 4. We can get this equitable by summarizing the numeral of leave above which have the hope figures of heads, and separate by the sum number of probably event

N #outcomes with N heads probability to get N heads 0 1 1/16 = 0.0625 1 4 4/16 = 0.25 3 6 6/16 = 0.375 4 4 4/16 = 0.25 5 1 1/16 = 0,0625

**Similar Questions**

**Question 1: If four coins are tossed, what is the probability of occurring neither 4 heads nor 4 tails?** **Solution:**

There can be 16 different probability when 4 coins are tossed : HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT There are 14 chances when we have neither 4 Heads nor 4 Tails.

Hence, the possibility or probability of occurring neither 4 Heads nor 4 Tails = 14/16 = 7/8

**Question 2: If four coins are tossed, find the possibility that there should be two heads and two tails.** **Solution:**

P ( A ) = { Number of favorable matter to A } ⁄ { sum issue of affair } By tossing four coins, the possibility are ( H, H, H, H ), ( T, H, H, H ), ( H, T, H, H ), ( H, H, T, H ), ( H, H, H, T ), ( T, T, H, H ), ( T, H, T, H ), ( T, H, H, T ), ( H, T, T, H ), ( H, T, H, T ), ( H, H, T, T ), ( T, T, T, H ), ( T, T, H, T ), ( T, H, T, T ), ( H, T, T, T ), ( T, T, T, T ) where H shows happening of head while tossing a coin and T shows happening of tail while tossing a mint. therefore, total count of likely affair = 16 hera, the favorable affair is occurring two heads and two tails on tossing four coins. intelligibly, the golden affair after tossing four coins are ( T, T, H, H ), ( T, H, T, H ), ( T, H, H, T ), ( H, T, T, H ), ( H, T, H, T ) and ( H, H, T, T ). consequently, Number of golden matter = 6 probability of occurring two heads and two tails =6/16=3/8. Hence, the possibility that there should be two heads and two tails after tossing four coins is 3/8 .

**Question 3: If you toss a coin 4 times, what is the probability of getting all heads?** **Solution:**

sample space : { ( HHHH ), ( HHHT ), ( HHTH ), ( HHTT ), ( HTHH ), ( HTHT ), ( HTTH ), ( HTTT ), ( THHH ), ( THHT ), ( THTH ), ( THTT ), ( TTHH ), ( TTHT ), ( TTTH ), ( TTTT ) } entire numeral of affairs = 16 Possibility getting all heads : P ( A ) = { Number of golden matter to A } ⁄ { total total of matter } = 1/16 i, HHHH

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