Suppose we roll two fair coins what is the probability of not getting all tails

I have now thought about this a bit more, and I think the way we get the information "one gets head" is important.

In the question, I put that the possible cases were:

• Head - Head
• Head - Tail
• Tail - Head
• Tail - Tail

Each tends to happen $25$% of the time.

If we see the first to fall comes up head, then we restrict to cases $1$ and $2$, and so the probabilities are $1/2$. Only in $50$% of the time we can see the first one coming up head, and in its half both coins end having head.

If the information was gotten from another person who sees the outcomes while you have not, and that person is forced to tell you that one coin came up head always that it happens, then that person will do it $75$% of the time, of which its $1/3$ is when both coins get head. So, once we know we are in those $75$%, we have $1/3$ probability.

If that another person was free to say you either "one coin came up head" or "one coin came up tail", then the probabilities are $1/2$ for both coins coming up head. The reason is that in the two cases we have got a head and a tail, that person has $1/2$ probability to report "one coin came up head".

$1)$ $25$% of the time we get two heads, and in all this cases that person tells you that one coin came up head.

$2)$ $50$% of the time we get a head and a tail, but only in half of its cases ($25$%), he tells you that one of them came up head. In the other half, he tells you that one of them came up tail.

$3)$ $25$% of the time we get two tails, and in all this cases that person tells you that one coin came up tail.

So, only in $25$% $+ 25$% $= 50$% he says you that one coin came up head, of which its half is when you have both coins head.

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.

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.

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.

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Problems on coin toss probability are explained here with different examples.When we flip a coin there is always a probability to get a head or a tail is 50 percent.

Suppose a coin tossed then we get two possible outcomes either a ‘head’ (H) or a ‘tail’ (T), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’.

The probability for equally likely outcomes in an event is:

Number of favourable outcomes ÷ Total number of possible outcomes

Total number of possible outcomes = 2

(i) If the favourable outcome is head (H).

Number of favourable outcomes = 1.

Therefore, P(getting a head)

= 1/2.

(ii) If the favourable outcome is tail (T).

Number of favourable outcomes = 1.

Therefore, P(getting a tail)

= 1/2.

Word Problems on Coin Toss Probability:

1. A coin is tossed twice at random. What is the probability of getting

(i) at least one head

(ii) the same face?

Solution:

The possible outcomes are HH, HT, TH, TT.

So, total number of outcomes = 4.

(i) Number of favourable outcomes for event E

                              = Number of outcomes having at least one head

                              = 3 (as HH, HT, TH are having at least one head).

So, by definition, P(F) = \(\frac{3}{4}\).

(ii) Number of favourable outcomes for event E

                              = Number of outcomes having the same face

                              = 2 (as HH, TT are have the same face).

So, by definition, P(F) = \(\frac{2}{4}\) = \(\frac{1}{2}\).

2. If three fair coins are tossed randomly 175 times and it is found that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and zero head appeared 35 times. 

What is the probability of getting 

(i) three heads, (ii) two heads, (iii) one head, (iv) 0 head. 

Solution: 

Total number of trials = 175. 

Number of times three heads appeared = 21. 

Number of times two heads appeared = 56. 

Number of times one head appeared = 63. 

Number of times zero head appeared = 35. 

(i) P(getting three heads)

= 21/175

= 0.12

(ii) P(getting two heads)

= 56/175

= 0.32

(iii) P(getting one head)

= 63/175

= 0.36

(iv) P(getting zero head)

= 35/175

= 0.20

Note: Remember when 3 coins are tossed randomly, the only possible outcomes

P(E1) + P(E2) + P(E3) + P(E4)

= (0.12 + 0.32 + 0.36 + 0.20)

= 1

3. Two coins are tossed randomly 120 times and it is found that two tails appeared 60 times, one tail appeared 48 times and no tail appeared 12 times.

If two coins are tossed at random, what is the probability of getting

(i) 2 tails,

(ii) 1 tail,

(iii) 0 tail

Solution:

Total number of trials = 120

Number of times 2 tails appear = 60 

Number of times 1 tail appears = 48

Number of times 0 tail appears = 12

(i) P(getting 2 tails)

= 60/120

= 0.50

(ii) P(getting 1 tail)

= 48/120

= 0.40

(iii) P(getting 0 tail)

= 12/120

= 0.10

Note:

P(E1) + P(E2) + P(E3)

= (0.50 + 0.40 + 0.10)

= 1

4. Suppose a fair coin is randomly tossed for 75 times and it is found that head turns up 45 times and tail 30 times. What is the probability of getting (i) a head and (ii) a tail?

Solution:

Total number of trials = 75.

Number of times head turns up = 45

Number of times tail turns up = 30

(i) Let X be the event of getting a head.

P(getting a head)

= 45/75

= 0.60

(ii) Let Y be the event of getting a tail.

P(getting a tail)

= 30/75

= 0.40

Note: Remember when a fair coin is tossed and then X and Y are the only possible outcomes, and

P(X) + P(Y)

= (0.60 + 0.40)

• 

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• 

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• 

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