Probability | Concepts | Problems

Probability | Concepts | Problems


Probability is the likelihood of an event to happen. Mathematically, probability of an event is given by, ((Number of outcomes in which the event can occur)/(Total number of outcomes))


 


Let us consider a coin toss. What is the probability of getting a heads?


There are two outcomes possible. (Heads and Tails) The required outcome (Heads) is only one. So, the probability is (1/2).

 


Now, let’s consider rolling of dice. When a dice is rolled, what is the probability of getting ‘1’ on the top surface?

Number of outcomes for getting ‘1’: 1

Total number of outcomes: 6 (1, 2, 3, 4, 5 and 6)

Probability = (1/6)

 


For the same situation, what is the probability of getting an even number?

Number of outcomes for getting an even number: 3 (2, 4 or 6)

Total number of outcomes: 6 (1, 2, 3, 4, 5 and 6)

Probability = (3/6) = (1/2)

 


Now, let’s consider tossing 2 coins. What is the probability that we get heads on both the coins?

Probability of getting heads on the first coin = (1/2)

Probability of getting heads on the second coin = (1/2)

 


Now, we need the probability of getting heads on the first coin AND the second coin.

NOTE: Whenever the situation has ‘and’, the probability is obtained by multiplying the two probabilities.

So, the probability of getting heads on both the coins is (1/2) * (1/2) = (1/4)

 


What is the probability of getting an odd number or ‘6’?

Outcomes for getting an odd number: 3 (1, 3 and 5)

Total outcomes: 6 (1, 2, 3, 4, 5 and 6)

Outcomes for getting ‘6’: 1

Total outcomes: 6 (1, 2, 3, 4, 5 and 6)Probability of getting an odd number = (3/6) = (1/2)

Probability of getting ‘6’ = (1/6)

We need the probability for getting an even number OR ‘6’

NOTE: Whenever the situation has ‘or’, the probability is obtained by adding the two probabilities.

So, probability of getting an odd number or ‘6’ = (1/2) + (1/6) = (4/6) = (2/3)

 


Q. What is the probability of getting an even number or a prime number?

Outcomes for getting an even number: 3 (2, 4, and 6)

Total number of outcomes: 6 (1, 2, 3, 4, 5, and 6)

Outcomes for getting a prime number: 3 (2, 3, and 5)

Total number of outcomes: 6 (1, 2, 3, 4, 5, and 6)

Probability of getting an even number = (3/6) = (1/2)

Probability of getting a prime number = (3/6) = (1/2)

But, the probability of getting a prime number or an even number is not [(1/2) + (1/2)]

Because getting probability as ‘1’ would mean the event will definitely happen. But, the event under consideration will not happen definitely. What if the outcome is ‘1’? ‘1’ is neither prime nor even.

To solve this problem, let’s understand why we got (1/2) for getting an even number and (1/2) for getting a prime number.

Even number [‘2’ OR ‘4’ OR ‘6’]

[(1/6){Probability of getting ‘2’} + (1/6){Probability of getting ‘4’} + (1/6){Probability of getting ‘6’}]

Prime number [‘2’ OR ‘4’ OR ‘6’]

[(1/6){Probability of getting ‘2’} + (1/6){Probability of getting ‘3’} + (1/6){Probability of getting ‘5’}]

We can see that the probability of getting ‘2’ has been accounted for, two times. So, we must remove one probability of getting a ‘2’.

So, the answer is [(1/6) + (1/6) + (1/6)] + [(1/6) + (1/6) + (1/6)] – (1/6) = (5/6)

 

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