If four coins are tossed, how many possible ways are there of at most 3 coins showing heads.

Hint- Here, we will be using the general formula for probability i.e., Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}$ in order to find the required probability.Given that we are tossing four coins and we have to find out the probability of obtaining two heads and two tails.According to general formula for probability of occurrence of an event, we can writeProbability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}$By tossing four coins, the possible outcomes 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 represents occurrence of head while tossing a coin and T represents occurrence of tail while tossing a coin.Therefore, Total number of possible outcomes = 16Here, the favourable event is getting two heads and two tails on tossing four coins.Clearly, the favourable outcomes 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).Therefore, Number of favourable outcomes = 6Probability of obtaining two heads and two tails $ = \dfrac{{\text{6}}}{{{\text{16}}}} = \dfrac{3}{8}$.Hence, the chance that there should be two heads and two tails after tossing four coins is $\dfrac{3}{8}$.Note- In these types of problems, where tossing of n coins is associated we already have a formula for calculating the total number of possible cases that will occur when n coins are tossed. i.e., Total number of possible outcomes when n coins are tossed =${2^{\text{n}}}$ (in this case n=4 that’s why total number of possible outcomes =${2^4} = 16$).

Probability is also known as the math of chance. This means the possibility, that deals in the occurrence of a likely affair. The value is deputed from zero to one. In math, Probability has been manifest to estimate how likely affairs are to occur. Basically, probability is the extent to which something is to be expected to occur.

Probability

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

Formula of Probability

P(A) = {Number of favourable affair to A} ⁄ {Total number of affair}

• 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.

Solution:

Each coin flip has 2 likely events, so the flipping of 4 coins has 2×2×2×2 = 16 likely events. We can summarize all likely events as follows, where H shows a head, 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 possibility 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 possibility for any one particular order of heads and tails is just (1/2)×(1/2)×(1/2)×(1/2)=(1/16).

Now lets ask: what is the possibility that in 4 flips, one gets N heads, where N = 0, 1, 2, 3, or 4. We can get this just by summarizing the number of result above which have the desired figures of heads, and dividing by the total number of likely 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 favourable affair to A } ⁄ {Total number 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 coin.

Therefore, Total number of likely affair = 16

Here, the favourable affair is occurring two heads and two tails on tossing four coins.

Clearly, the favourable 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).

Therefore, Number of favourable affair = 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)}

Total number of affairs = 16  

Possibility getting all heads :

P(A) = {Number of favourable affair to A} ⁄ {Total number of affair}

= 1/16  i.e., HHHH

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Is there a way to solve the problem considering that the probability of getting a head is 1/2 and then calculating $.5^4$ and multiplying $.5^4$ by 4 as there are 4 ways that this could occur?