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Probability

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Probability Lessons

Playing Cards Nomenclature & Probability
In a pack or deck of 52 playing cards, they are divided into 4 suits of 13 cards each i.e. spades ♠ hearts ♥, diamonds ♦, clubs ♣. Total Blck Cards = 26 Total Red Cards...

Bernouli Distributiions
Bernoulli Distribution: Let’s start with the easiest distribution that is Bernoulli Distribution. It is actually easier to understand than it sounds! All you cricket junkies out there! At the...

Uniform Distribution
Uniform Distribution: When you roll a fair die, the outcomes are 1 to 6. The probabilities of getting these outcomes are equally likely and that is the basis of a uniform distribution. Unlike Bernoulli...

Binomial Distribution
Binomial Distribution: Let’s get back to cricket. Suppose that you won the toss today and this indicates a successful event. You toss again but you lost this time. If you win a toss today, this...

Normal Distribution
Normal distribution represents the behavior of most of the situations in the universe (That is why it’s called a “normal” distribution. I guess!). The large sum of (small) random variables...

Probability Questions

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Answered on 02/12/2016 Learn Probability +2 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Sreedevi Addepalli

Teacher

The probability of the outcome of an experiment is never negative.
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Answered on 02/12/2016 Learn Probability +2 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Sreedevi Addepalli

Teacher

No, because the number of successes and failures should be equal to 1.
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Answered on 02/12/2016 Learn Probability +2 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Sreedevi Addepalli

Teacher

In the single throw of a die, the sample space n(S) = 6; let 'E' be the event of getting '5'. n(E) = 1; P(E) = n(E) /n(S); P(E) = 1/6.
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Answered on 02/12/2016 Learn Probability +2 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Sreedevi Addepalli

Teacher

There will be 4 aces (Event E) that belongs to 4 different types say hearts, clubs, diamonds and spades. The probability of getting an ace of 52 cards (sample space = 52) is given by the ratio of favourable events to sample space. P(E) = n(E) / n(S); P(E) = 4/52; P(E) = 1/4.
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