Probability and Statistics Lecture 1

Basic Probability

Basic Probability

• Sample Space: The set of all possible outcomes of an experiment is known as the sample space, denoted by $S$.
• Event: A particular set of outcomes from the sample space, denoted by $A$.

Laws of probabilities:

1. $0\le P(A)\le 1$
2. $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
3. $P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$
4. If $A$ and $B$ are exclusive events, i.e., disjoint events, then $P(A\cup B)=P(A)+P(B)$
5. $P(A^C)=1-P(A)$

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Example 1: If two dice are rolled, what is the probability that the sum of the upturned faces will equal 7? Solution: Let $n(S)=36$. Let $A$ be the event that the sum is 7, so $A=\{(1,6),(6,1),(3,4),(4,3),(2,5),(5,2)\}$. Therefore, $n(A)=6$. Thus, $P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}.$

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Example 2: A fair coin is tossed 4 times. Define the sample space corresponding to this random experiment. Also, give the subsets corresponding to the following events and find the respective probabilities:

Solution: Let $S$ be the sample space: $S=\{HHHH,HHTH,HHHT,HHTT,HTTH,HTTT,THHH,THTH,THTT,TTHH,TTHT,TTTT\}$

(a) More heads than tails are obtained:

Let $A=\{HHHH,HHTH,HHHT,HHTT\}$.

Thus, $P(\text{more heads than tails})=\frac{5}{16}.$

(b) Tails occur on all even-numbered tosses:

Let $B=\{HTHT,HTHT,TTHH,TTTT\}$.

Thus, $P(B)=\frac{4}{16}=\frac{1}{4}.$

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Example 3: If 3 balls are randomly drawn from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and the other two are black?

Solution: The total number of balls is 11. Therefore, $n(S)=\left(\genfrac{}{}{0ex}{}{11}{3}\right).$

We want to select 1 white ball and 2 black balls. The number of favorable outcomes is $\left(\genfrac{}{}{0ex}{}{6}{1}\right)\times \left(\genfrac{}{}{0ex}{}{5}{2}\right).$

Thus, the probability is $P(A)=\frac{\left(\genfrac{}{}{0ex}{}{6}{1}\right)\times \left(\genfrac{}{}{0ex}{}{5}{2}\right)}{\left(\genfrac{}{}{0ex}{}{11}{3}\right)}=\frac{4}{11}.$

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Example 4: From 6 positive and 8 negative numbers, 4 numbers are chosen at random and multiplied. What is the probability that the product is positive?

Solution: We can select 4 numbers in the following cases:

• Case 1: All are positive.
• Case 2: Two positive and two negative.
• Case 3: All are negative.

The total number of ways to select 4 numbers is $n(S)=\left(\genfrac{}{}{0ex}{}{14}{4}\right).$

The number of ways the product is positive is $6C4+6C2\cdot 8C2+8C4.$

Thus, the probability is $P(A)=\frac{6C4+6C2\cdot 8C2+8C4}{n(S)}=\frac{505}{1001}.$

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References

• Lectures
• Teams Notes

Continued

Probability and Statistics Lecture 2

Information

• date: 2025.02.03
• time: 11:23