# If a dice is rolled 5 times, what is the probability of rolling a number less than 3 at least 3 times?

In simple words, the probability is the study of the possibility of occurrence of a particular event with respect to a number of other possible events, among which no two or more of them can occur simultaneously, that is at a given point in time only one of the possible events can occur. The simplest example of the application of probability is to determine the possibility of occurrence of a head that is ½, viz. favorable outcomes (occurrence of head) divided by possible outcomes (head and tail).

### Binomial distribution

Let’s proceed with the same activity of tossing a coin, now suppose your friend suggests you throw a coin 3 times, and if a head appears at least once you have to throw him a treat. But, you know that you are left with very little money to spend. Here it becomes necessary for you to get an idea as to what would be the probability of you being forced to throw a treat.

In such cases, where success and failure are involved in independent trials, the process used to calculate the probability is known as the binomial distribution method or the Bernoulli Distribution method.

It has got such a name because it involves only two possible events that are success and failure that are distributed throughout the trials executed.

- In order to use this method certain points should be kept in mind:
- There should be a fixed and finite number of trials.
- In each trial, the probability of success and hence failure should be constant.
- Each trial should be independent of the other trials.

These points if fulfilled together, allow the usage of the Binomial Distribution method. Now getting back to the conditions set by your friend, we can see that all the required conditions to apply the binomial distribution method are fulfilled. This was a practical problem, now let’s work out and solve a theoretical problem on binomial distribution to understand it better.

Let’s suppose there is an unbiased six-faced dice and to find the probability of rolling a number less than 3 at least 3 times within a total of 5 rolls of the dice.

**Basic Principle of arrangement**

For a better understanding of questions on the binomial distribution, it is important to understand the concept of arranging things in a row. There may be a number of ways to arrange identical or non-identical things, sometimes counting these arrangements becomes important. If there are (n + m) things out of which n are identical of one kind and m are identical of some other kind, then the total ways of arranging them in a row are given by (n + m)!/(m! × n!).

### If a dice is rolled 5 times, what is the probability of rolling a number less than 3 at least 3 times?

**Solution:**

Steps to proceed

- According to the question, our favorable outcome is ⇢ Among the 5 independent rolls, there maybe 3 or 4 or all 5 numbers appearing less than 3.
- Consider the favorable outcome as the success event(S) and calculate its probability, i.e. P(S) = 2/6. Hence out failure event(F) has a probability of, P(F)= 1 – P(S) = 4/6.
- Now just to draw a similarity with the Basic principle of arrangement concept, that we will be using soon, let us consider the probability of success as a thing of one kind and the probability of failure P(F) as a thing of another kind non-identical from P(S). There may be three possible cases:

- Event E
_{1 }– 3 numbers less than 3 and 2 numbers greater than or equal to 3. P(E_{1}) = (5!/(3! × 2!)) × (P(S))^{3}× (P(F))^{2}= (5!/(3! × 2!)) × (1/3)

^{3}× (2/3)^{2}= 40/243

- Event E
_{2}– 4 numbers less than 3 and 1 number greater than or equal to 3. P(E_{2}) = (5!/(4! × 1!)) × (P(S))^{4}× (P(F))= (5!/(4! × 1!)) × (1/3)

^{4}× (2/3)= 10/243

- Event E
_{3}– 5 numbers less than 3. p(E_{3}) = (5!/(5! × 0!)) × (P(S))^{5}× (P(F))^{0 }= (1/3))

^{5}× (2/3)^{0}= 1/243

Use the formula of the basic principle of combination because, here all successful events can be considered as identical and the failure events as identical but of a different kind, hence the number of ways of their arrangement in a row can be counted by using the basic principle of combination. For the final answer we need the union of all the three possibilities, i.e., (a) or (b) or (c). Let the final probability be P(Q) = (a) + (b) + (c)

= 40/243 + 10/243 + 1/243

= 51/243

**Sample Examples**

**Question1: A coin is tossed 3 times, what is the probability that the tail will appear only once?**

**Solution:**

Let the probability of getting a head be P(H) = 1/2 and that of getting a tail to be P(T) = 1/2

Let the event of our favorable outcome, that is tail will appear only once out of the three performed trials be P(E)

All the conditions to apply the Binomial Distribution formula are fulfilled, therefore,

P(E) = 3!/(1! × 2!) × (P(H))

^{2}× (P(T)); if the tail has to appear only once the head has to appear twice.= 3!/(1! × 2!) × (1/2))

^{2}× (1/2)= 3/8

**Question2: A person can hit a bullseye with a probability of 20%. What is the probability that he can hit it at most 3 times if 4 trials are given?**

**Solution:**

Let the probability of hitting the target be P(H) = 20/100 = 1/5 and that of failing to hit it be P(FH) = 1 – P(H)

= 4/5

Our favorable outcome is to get 3, 2 ,1 or no hits at all in the four trials, let it be P(F)

All conditions to apply Binomial Distribution formula are fulfilled, therefore;

If he fails to hit every time then; P1 = 4!/4! × (P(FH))

^{4}= (4/5)

^{4}= 256/625

If he hits it once; P

_{2}= 4!/(1! × 3!) × (P(H)) × (P(FH))^{3}= 4!/(1! × 3!) × (1/5) × (4/5))

^{3}= 256/625

If he hits it twice; P

_{3}= 4!/(2! × 2!) × (P(H))^{2}× (P(FH))^{2}= 4!/(2! × 2!) × (1/5)

^{2}× (4/5)^{2}= 96/625

If he hits it thrice; P

_{4 }= 4!/(3! × 1!) × (P(H))^{3}× (P(FH))= 4!/(3! × 1!) × (1/5)

^{3}× (4/5)= 16/625

Therefore, P(F) = P

_{1}+ P_{2}+ P_{3}+ P_{4}

= 624/625

**Question3: Two friends Rakesh and Simi were throwing unbiased dice one after the other. They have decided to stop only when one of them got an even number. Can the probability of Simi winning the game be obtained by the Binomial Distribution method? Why/Why not?**

**Solution:**

Let the probability of getting an odd number on an unbiased dice be P(O) = 1/2 and that of getting an even number be P(E) = 1/2

Verifying required conditions:

- Each trial by either friend is independent of all other trials
- Only two events are possible one is the success and the other failure
- With each trial the probability of success and failure does not change
- There may be an infinite number of trials
The 4th point violates the conditions required to apply the Binomial Distribution method, because it may be possible that an even number appeared at the nth trial where n tends to infinity.

Hence the question cannot be solved by the Binomial Distribution method.

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