Probability Calculations

Assume there was an experiment where a dice was rolled 500 times, and you see 74 fives. The experiential probability of throwing a five is \(\frac{74}{500} = 0.148\).

Assuming the die is fair, then the theoretical probability of throwing a 5 is \(\frac{1}{6} = 0.166\).

This is because the 6 is the total possible outcomes there could be from throwing a dice. And 1 indicates the desired outcome (rolling a five).

So on average, if the die is fair, you will see \(\frac{1}{6} \space of \space 500 = 83\) fives.

This mathematics means that by throwing the dice 500 times, provided the dice is fair, you're going to roll approximately 83 fives.

When two coins are tossed simultaneously, what is the probability of getting heads on one coin and tails on the other?

Answer:

The sample space for when two coins are tossed is as follows;

S = {HH,HT,TH,TT}

Where H= Heads and T= Tails

\(Probability \space of \space an \space event = \frac {Number \space of \space favourable \space outcomes}{Total \space number \space of \space outcomes}\)

\(\begin{align}P(Head \space on \space one \space and \space Tail \space and \space on \space other) &= P(HT) + P(TH) \\ &= \frac {1}{4} + \frac {1}{4} \\ &= \frac {2}{4} \end{align}\)

Divide through by 2

\(P(Head \space on \space one \space and \space Tail \space on \space the \space other) = \frac {1}{2}\)

This means there is a 50% chance of having a head on one and a tail on another after tossing coins simultaneously.

What is the probability that a random card drawn from a pack of cards is a face card?

answer:

Since a standard deck has 52 cards, the total number of outcomes will be 52.

n(S) = 52

Now let E be the event of drawing a face card.

Number of favourable events = n (E)

\(n(E) = 3 \cdot 4 = 12\)

\(Probability = \frac{Number \space of \space Favourable \space Outcomes}{Total \space Number \space of \space Outcomes}\)

\(\begin{align} P(E) &= \frac{n(E)}{n(S)} \\ &= \frac {12}{52} \\ &=\frac{3}{13} \\ &=0.23 \end{align}\)

This can also be expressed in percentages as 23% probability.

We have a box that contains 4 blue balls, 5 red balls, and 11 white balls. If three balls get drawn from the vessel at random, what is the probability that the first ball is red, the second ball is blue, and the third ball is white?

Answer:

We will first find the probability for each colour picked.

Since there are 20 balls in all, the possible outcome for a pick is 20.

The probability that the first ball will be red then is \(\frac{5}{20}\).

We now have picked a ball which leaves us with the possible outcomes being 19.

So the probability of the second pick being blue is \(\frac{4}{19}\).

Again, because we have another pick already being made, our total possible outcomes reduce by 1, leaving us with 18.

The probability that the third ball is white is \(\frac{11}{18}\).

Therefore the entire probability that the first ball is red, the second ball is blue, and the third ball is white is \(\frac{5}{20} \cdot \frac {4}{19} \cdot \frac {11}{18} = \frac {44}{1368}\)

P = 0.032

This can also be expressed in percentages as 3.2% probability.