Two fair coins are tossed simultaneously a what is the probability of getting only one head

You can understand probability by thinking about flipping a coin.

Probability is a field of mathematics that deals with calculating the likelihood of occurrence of a specific event.

The likelihood of an event is expressed as a number between zero (the event will never occur) and one (the event is certain). For example, the probability of an outcome of heads on the toss of a fair coin is ½ or 0.5. The probability of an event can also be expressed as a percentage (e.g., an outcome of heads on the toss of a fair coin is 50% likely) or as odds (e.g., the odds of heads on the toss of a fair coin is 1:1).

A single toss of a coin is an event (also called a trial) that is not connected to or influenced by other events. When a coin is tossed twice, the coin has no memory of whether it came up heads or tails the first time, so the second toss of the coin is independent. The probability of heads on the first toss is 50%, just as it is on all subsequent tosses of the coin.

The two outcomes of the toss of a coin are heads or tails. For any individual toss of the coin, the outcome will be either heads or tails. The two outcomes (heads or tails) are therefore mututally exclusive; if the coin comes up heads on a single toss, it cannot come up tails on the same toss.

There are two useful rules for calculating the probability of events more complicated than a single coin toss.

The first is the Product Rule. This states that the probability of the occurrence of two independent events is the product of their individual probabilities. The probability of getting two heads on two coin tosses is 0.5 x 0.5 or 0.25.

A visual representation of the toss of two coins.

The Product Rule is evident from the visual representation of all possible outcomes of tossing two coins shown above. The probability of getting heads on the toss of a coin is 0.5. If we consider all possible outcomes of the toss of two coins as shown, there is only one outcome of the four in which both coins have come up heads, so the probability of getting heads on both coins is 0.25.

The second useful rule is the Sum Rule. This states that the probability of the occurrence of two mutually exclusive events is the sum of their individual probabilities. As you can see from the picture, the probability of getting one head and one tail on the toss of two coins is 0.5. There are two different ways that this can happen. The first coin can come up heads and the second coin can come up tails, or the first coin can come up tails and the second coin can come up heads. In any single trial, it is not possible for both of these outcomes to occur, so these are mutually exclusive.

There are four possible mutually exclusive outcomes on the toss of two coins as shown, each with a probability of 0.25. The sum of the probability of two of these outcomes (heads, tails or tails, heads) is 0.25 + 0.25 or 0.5.

Probability applies to breeding horses as well as tossing coins.

The basic rules of probability apply to horse breeding as well. Horses have two copies of each of their genes. A horse that is heterozygous for the mutation that causes HYPP has the genotype n/H. When this horse makes gametes (sperm or egg), there is only one copy of each gene in the gamete. There is a 50% chance that a gamete has the n allele and a 50% chance that a gamete has the H allele.

The process of fertilization is like the toss of two coins. If a stallion that is n/H is bred to a mare that is n/H, the chance that the foal will be n/n is 0.25, while the chance that the foal will be n/H is 0.5. The Sum Rule and the Product Rule apply to horse breeding in the same way that they apply to coin tosses.

BUY TEST

Contact Us

Connect with Us

EquiSeq is engaged in ongoing research to develop genetic tests for inherited diseases in horses.

We need your financial contributions to accelerate our research. Our researchers need laboratory supplies to isolate and analyze DNA samples to identify disease genes.

Donate $25 USD today!

Donate

We make it easy. Pay by credit card, debit card, or use your PayPal account.

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

Let us take the experiment of tossing two coins simultaneously:

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 22 = 4

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

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

1. Two different coins are tossed randomly. Find the probability of:

(i) getting two heads

(ii) getting two tails

(iii) getting one tail

(iv) getting no head

(v) getting no tail

(vi) getting at least 1 head

(vii) getting at least 1 tail

(viii) getting atmost 1 tail

(ix) getting 1 head and 1 tail

Solution:

When two different coins are tossed randomly, the sample space is given by

S = {HH, HT, TH, TT}

Therefore, n(S) = 4.

(i) getting two heads:

Let E1 = event of getting 2 heads. Then,
E1 = {HH} and, therefore, n(E1) = 1.
Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4.

(ii) getting two tails:

Let E2 = event of getting 2 tails. Then,
E2 = {TT} and, therefore, n(E2) = 1.
Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4.

(iii) getting one tail:

Let E3 = event of getting 1 tail. Then,
E3 = {TH, HT} and, therefore, n(E3) = 2.
Therefore, P(getting 1 tail) = P(E3) = n(E3)/n(S) = 2/4 = 1/2

(iv) getting no head:

Let E4 = event of getting no head. Then,
E4 = {TT} and, therefore, n(E4) = 1.
Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = ¼.

(v) getting no tail:

Let E5 = event of getting no tail. Then,
E5 = {HH} and, therefore, n(E5) = 1.
Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼.

(vi) getting at least 1 head:

Let E6 = event of getting at least 1 head. Then,
E6 = {HT, TH, HH} and, therefore, n(E6) = 3.
Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.

(vii) getting at least 1 tail:

Let E7 = event of getting at least 1 tail. Then,
E7 = {TH, HT, TT} and, therefore, n(E7) = 3.
Therefore, P(getting at least 1 tail) = P(E2) = n(E2)/n(S) = ¾.

(viii) getting atmost 1 tail:

Let E8 = event of getting atmost 1 tail. Then,
E8 = {TH, HT, HH} and, therefore, n(E8) = 3.
Therefore, P(getting atmost 1 tail) = P(E8) = n(E8)/n(S) = ¾.

(ix) getting 1 head and 1 tail:

Let E9 = event of getting 1 head and 1 tail. Then,
E9 = {HT, TH } and, therefore, n(E9) = 2.
Therefore, P(getting 1 head and 1 tail) = P(E9) = n(E9)/n(S)= 2/4 = 1/2.

The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins.

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability