When 4 fair coins are tossed together what is the probability of getting at least three heads 3 4 o 5/16 O 5 8?

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?

Here we will learn how to find the probability of tossing three coins.

Let us take the experiment of tossing three coins simultaneously:

When we toss three coins simultaneously then the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 23 = 8

The above explanation will help us to solve the problems on finding the probability of tossing three coins.

Worked-out problems on probability involving tossing or throwing or flipping three coins:

1. When 3 coins are tossed randomly 250 times and it is found that three heads appeared 70 times, two heads appeared 55 times, one head appeared 75 times and no head appeared 50 times.

If three coins are tossed simultaneously at random, find the probability of:

(i) getting three heads,

(ii) getting two heads,

(iii) getting one head,

(iv) getting no head

Solution:

Total number of trials = 250.

Number of times three heads appeared = 70.

Number of times two heads appeared = 55.

Number of times one head appeared = 75.

Number of times no head appeared = 50.

In a random toss of 3 coins, let E1, E2, E3 and E4 be the events of getting three heads, two heads, one head and 0 head respectively. Then,

(i) getting three heads

P(getting three heads) = P(E1) Number of times three heads appeared

= Total number of trials

= 70/250

= 0.28

(ii) getting two heads

P(getting two heads) = P(E2) Number of times two heads appeared

= Total number of trials

= 55/250

= 0.22

(iii) getting one head

P(getting one head) = P(E3) Number of times one head appeared

= Total number of trials

= 75/250

= 0.30

(iv) getting no head

P(getting no head) = P(E4) Number of times on head appeared

= Total number of trials

= 50/250

= 0.20

Note:

In tossing 3 coins simultaneously, the only possible outcomes are E1, E2, E3, E4 and P(E1) + P(E2) + P(E3) + P(E4)

= (0.28 + 0.22 + 0.30 + 0.20)

= 1

2. When 3 unbiased coins are tossed once.

What is the probability of:

(i) getting all heads

(ii) getting two heads

(iii) getting one head

(iv) getting at least 1 head

(v) getting at least 2 heads

(vi) getting atmost 2 heads

Solution:

In tossing three coins, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

And, therefore, n(S) = 8.

(i) getting all heads

Let E1 = event of getting all heads. Then,
E1 = {HHH}
and, therefore, n(E1) = 1.
Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

(ii) getting two heads

Let E2 = event of getting 2 heads. Then,
E2 = {HHT, HTH, THH}
and, therefore, n(E2) = 3.
Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.

(iii) getting one head

Let E3 = event of getting 1 head. Then,
E3 = {HTT, THT, TTH} and, therefore,
n(E3) = 3.
Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

(iv) getting at least 1 head

Let E4 = event of getting at least 1 head. Then,
E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}
and, therefore, n(E4) = 7.
Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) getting at least 2 heads

Let E5 = event of getting at least 2 heads. Then,
E5 = {HHT, HTH, THH, HHH}
and, therefore, n(E5) = 4.
Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.

(vi) getting atmost 2 heads

Let E6 = event of getting atmost 2 heads. Then,
E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}
and, therefore, n(E6) = 7.
Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8

3. Three coins are tossed simultaneously 250 times and the outcomes are recorded as given below.

Outcomes

3 heads

2 heads

1 head

No head

Total

Frequencies

100

250

If the three coins are again tossed simultaneously at random, find the probability of getting

(i) 1 head

(ii) 2 heads and 1 tail

(iii) All tails

Solution:

(i) Total number of trials = 250.

Number of times 1 head appears = 100.

Therefore, the probability of getting 1 head

= $\frac{\textrm{Frequency of Favourable Trials}}{\textrm{Total Number of Trials}}$

= $\frac{\textrm{Number of Times 1 Head Appears}}{\textrm{Total Number of Trials}}$

= $\frac{100}{250}$

= $\frac{2}{5}$

(ii) Total number of trials = 250.

Number of times 2 heads and 1 tail appears = 64.

[Since, three coins are tossed. So, when there are 2 heads, there will be 1 tail also].

Therefore, the probability of getting 2 heads and 1 tail

= $\frac{\textrm{Number of Times 2 Heads and 1 Trial appears}}{\textrm{Total Number of Trials}}$

= $\frac{64}{250}$

= $\frac{32}{125}$

(iii) Total number of trials = 250.

Number of times all tails appear, that is, no head appears = 38.

Therefore, the probability of getting all tails

= $\frac{\textrm{Number of Times No Head Appears}}{\textrm{Total Number of Trials}}$

= $\frac{38}{250}$

= $\frac{19}{125}$.

These examples will help us to solve different types of problems based on probability of tossing three coins.

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