8th Grade - Compound Probability

Introduction

• Probability is a measure of the likelihood of the occurrence of an event.
• Mathematically, the probability of an event A is calculated using the formula: $P\left(A\right)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
• Probability is of two types - theoretical probability & experimental probability.
• Theoretical probability is what we expect to happen in an experiment (remains the same), whereas experimental probability is what actually happens when we try it out (not always the same).
• The probability of an event ranges from 0 to 1.
• Probability can be expressed in terms of fractions, percentages, or decimals.
• The probability related to the occurrence of two or more events is called Compound probability.
• A combination of such events is called compound events.

Independent and Dependent Events

• Compound events can be of two types - independent & dependent.
• Independent Events:

• • Events that don't depend on each other (the outcome of one event does not affect the outcome of the second event) are called Independent events.
  • Example: getting tails in a coin flip and rolling five on a six-sided die
  • The probability of two independent events, A and B, can be found using the formula below.

$P\left(A \text{and} B\right)\mathit{=}P\left(A\right)\mathit{·}P\left(B\right)$

• • Thus to find the compound probability of two independent events, we simply multiply their individual probabilities.

• Dependent Events:

• • Events that depend upon each other are called Dependent events.
  • Example: drawing two cards from a standard without replacement
  • The probability of two independent events, A and B, can be found using the formula below.

$P\mathit{(}A\mathit{ }\mathit{\text{and}}\mathit{ }B\mathit{)}\mathit{=}P\mathit{\left(}A\mathit{\right)}\mathit{·}P\mathit{\left(}B\right|A\mathit{)}$

• • Thus to find the compound probability of two dependent events, we multiply the probability of the first event $P\left(A\right)$ by the probability of the next event after the event has taken place $P\left(B|A\right)$.

Mutually Exclusive and Mutually Inclusive Events

• Events can also be categorized as mutually exclusive and mutually inclusive.
• Mutually Exclusive Events:

• • Compound events that cannot happen at the same time are called mutually exclusive events.
  • Example: getting a number 4 and an odd number when rolling a six-sided die once
  • For two mutually exclusive events, A and B,

$\mapsto P\left(A \text{and }B\right)=0$ because A and B cannot happen at the same time.

$\mapsto P\left(A \text{or} B\right)=P\left(A\right)+P\left(B\right)$

• Mutually Inclusive Events:

• • Two events that can happen at the same time simultaneously are called mutually inclusive events.
  • Example: getting a number less than four and an odd number when rolling a six-sided die once
  • For two mutually inclusive events, A and B,

$\mapsto P\left(A \text{or} B\right)=P\left(A\right)+P\left(B\right)-P\left(A \text{and} B\right)$

• Note: When dealing with compound probabilities, we often encounter two symbols. The symbol '$\cup$' means "the union of" which is equivalent to $P\left(A \text{or} B\right)$ and '$\cap$' means "the intersection of" which is equivalent to $P\left(A \text{and }B\right)$.

Solved Examples

Question 1: A card is drawn randomly from a well-shuffled deck of 52 cards. Find the probability that the card drawn is a queen or an ace.

Solution: Drawing a queen and an ace card are mutually exclusive events, as they cannot happen simultaneously.

$P\left(A\right)=P\left(\text{drawing a queen}\right)=\frac{13}{52}=\frac{1}{4}$

$P\left(B\right)=P\left(\text{drawing an ace card}\right)=\frac{4}{52}=\frac{1}{13}$

$P{\left(A \text{or }B\right)}_{mutually exclusive}=P\left(A\right)+P\left(B\right)$

$=\frac{1}{4}+\frac{1}{13}$

$=\frac{13+4}{52}$

$=\frac{17}{52}$

Question 2:  A bowl contains ten green marbles, six red marbles, and four black marbles. Find the probability of drawing a green marble and then a red marble.

Solution: Total number of balls = 10 + 6 + 4 = 20

Probability of drawing green ball first, $P\left(\text{green}\right)=P\left(A\right)=\frac{10}{20}=\frac{1}{2}$

Probability of drawing a red ball after the green ball is taken out, $P\left(\text{red}\right)=P\left(B\right)=\frac{6}{19}$ Note that after drawing a green ball, 19 ball remains.

Required probability, $P\left(A \text{and} B\right)=P\left(A\right)·P\left(A|B\right)$$=\frac{1}{2}\times \frac{6}{19}=\frac{3}{19}$

Cheat Sheet

• $P\left(\text{event}\right)=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
• If A and B are two independent events, then $P\left(A \text{and }B\right)=P\left(A\right)·P\left(B\right)$.
• If A and B are two dependent events, then $P\left(A \text{and }B\right)=P\left(A\right)·P\left(B|A\right)$.
• If A and B are two mutually exclusive events, then $P\left(A \text{and }B\right)=0$. Also, $P\left(A orB\right)=P\left(A\right)+P\left(B\right)$.
• If A and B are two mutually inclusive events, then $P\left(A orB\right)=P\left(A\right)+P\left(B\right)-P\left(A \text{and} B\right)$.

Blunder Areas

• Sometimes a case may arise where there are a lot of favorable outcomes and fewer non-favorable outcomes. In such cases, we should first find the probability of non-favorable outcomes and then subtract it from the total number of outcomes.