Probability Concepts

1. Basic Probability

• Formula:

$$P(E)=\frac{\text{Number of favorable outcomes}}{\text{Total number of equally likely outcomes}}$$

Also:

$$0\le P(E)\le1$$

• Example: A die is rolled once. Find the probability of getting an even number. Solution: Possible outcomes:

$${1,2,3,4,5,6}$$

Favorable outcomes:

$${2,4,6}$$

Therefore:

$$P(E)=\frac36=\boxed{\frac12}$$

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2. Complementary Events

• Formula:

$$P(\text{not }A)=1-P(A)$$

Hence:

$$P(A)=1-P(\text{not }A)$$

• Example: A die is rolled. Find the probability of not getting a 6. Solution:

$$P(6)=\frac16$$

Therefore:

$$P(\text{not }6)=1-\frac16=\boxed{\frac56}$$

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3. Independent Events

• Formula: For independent events:

$$P(A\cap B)=P(A)P(B)$$

The occurrence of one event does not affect the probability of the other.

• Example: A coin is tossed and a die is rolled. Find the probability of getting a head and a 6. Solution:

$$P(H)=\frac12,\qquad P(6)=\frac16$$

Since the events are independent:

$$P(H\cap6)=\frac12\times\frac16 =\boxed{\frac1{12}}$$

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4. Mutually Exclusive Events

• Formula: If A and B cannot occur together:

$$P(A\cap B)=0$$

Therefore:

$$P(A\cup B)=P(A)+P(B)$$

• Example: A die is rolled. Find the probability of getting either 2 or 5. Solution: The events are mutually exclusive.

$$P(2)=\frac16,\qquad P(5)=\frac16$$

Therefore:

$$P(2\text{ or }5)=\frac16+\frac16=\boxed{\frac13}$$

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5. Union of Two Events

• Formula:

$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$

The intersection must be subtracted because it is counted twice.

• Example: A card is drawn from a standard deck. Find the probability of getting a king or a heart. Solution:

$$P(K)=\frac4{52}$$ $$P(H)=\frac{13}{52}$$

The king of hearts belongs to both:

$$P(K\cap H)=\frac1{52}$$

Therefore:

$$P(K\cup H) =\frac4{52}+\frac{13}{52}-\frac1{52}$$ $$=\boxed{\frac4{13}}$$

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6. Conditional Probability

• Formula:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

For equally likely outcomes:

$$P(A\mid B)=\frac{n(A\cap B)}{n(B)}$$

• Example: A card is drawn from a deck. Given that the card is a face card, find the probability that it is a king. Solution: There are 12 face cards:

$$J,Q,K\text{ of 4 suits}$$

There are 4 kings.

Therefore:

$$P(K\mid\text{face card}) =\frac4{12} =\boxed{\frac13}$$

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7. Cards Drawn Without Replacement

• Formula: When objects are drawn without replacement, the total number of objects changes after each draw:

$$P(A\text{ then }B)=P(A)\times P(B\mid A)$$

• Example: A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red. Solution: First red:

$$\frac58$$

After one red is removed:

$$P(\text{second red}\mid\text{first red})=\frac47$$

Therefore:

$$P(\text{both red}) =\frac58\times\frac47 =\boxed{\frac5{14}}$$

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8. Cards Drawn With Replacement

• Formula: When the object is replaced after each draw, the probabilities remain unchanged:

$$P(A\text{ then }B)=P(A)P(B)$$

• Example: A card is drawn from a deck and replaced. Another card is drawn. Find the probability that both cards are aces. Solution: Probability of an ace on each draw:

$$\frac4{52}=\frac1{13}$$

Since the card is replaced:

$$P(\text{both aces}) =\frac1{13}\times\frac1{13} =\boxed{\frac1{169}}$$

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9. Probability of At Least One Occurrence

• Formula:

$$P(\text{at least one})=1-P(\text{none})$$

For (n) independent trials with success probability (p):

$$P(\text{at least one success})=1-(1-p)^n$$

• Example: A coin is tossed 3 times. Find the probability of getting at least one head. Solution: Probability of no heads = all tails:

$$P(TTT)=\left(\frac12\right)^3=\frac18$$

Therefore:

$$P(\text{at least one H}) =1-\frac18 =\boxed{\frac78}$$

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10. Exactly (r) Successes in (n) Trials

• Formula: For independent trials:

$$P(X=r)={}^{n}C_rp^r(1-p)^{n-r}$$

• Example: A coin is tossed 4 times. Find the probability of getting exactly 2 heads. Solution:

$$n=4,\quad r=2,\quad p=\frac12$$

Therefore:

$$P(X=2) ={}^{4}C_2\left(\frac12\right)^2 \left(\frac12\right)^2$$ $$=6\times\frac1{16} =\boxed{\frac38}$$

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11. Bayes’ Theorem

• Formula:

$$P(A_i\mid B) ============ \frac{P(A_i)P(B\mid A_i)} {\sum_jP(A_j)P(B\mid A_j)}$$

For two cases:

$$P(A\mid B)= \frac{P(A)P(B\mid A)} {P(A)P(B\mid A)+P(A')P(B\mid A')}$$

• Example: Urn I has 3 red and 2 blue balls. Urn II has 4 red and 1 blue ball. A die is rolled: even means Urn I, odd means Urn II. A red ball is drawn. Find the probability it came from Urn I. Solution:

$$P(I)=P(II)=\frac12$$ $$P(R\mid I)=\frac35,\qquad P(R\mid II)=\frac45$$

Therefore:

$$P(I\mid R) = \frac{\frac12\cdot\frac35} {\frac12\cdot\frac35+\frac12\cdot\frac45}$$ $$=\frac{3/10}{7/10} =\boxed{\frac37}$$

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12. Two-Stage Urn / Bag Probability

• Formula: For alternative sources:

$$P(E)=P(A)P(E\mid A)+P(B)P(E\mid B)$$

This is the law of total probability.

• Example: Bag A contains 3 red and 2 blue balls. Bag B contains 2 red and 3 blue balls. A bag is selected randomly and one ball is drawn. Find the probability of getting a red ball. Solution:

$$P(A)=P(B)=\frac12$$ $$P(R\mid A)=\frac35,\qquad P(R\mid B)=\frac25$$

Thus:

$$P(R)=\frac12\cdot\frac35+\frac12\cdot\frac25$$ $$=\frac3{10}+\frac2{10} =\boxed{\frac12}$$

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13. Geometric Probability — Length

• Formula:

$$P=\frac{\text{Favorable length}}{\text{Total length}}$$

• Example: A point is chosen randomly on a stick of length 1 m. Find the probability that it lies within 20 cm of the midpoint. Solution: Total length:

$$100\text{ cm}$$

Favorable region:

$$20+20=40\text{ cm}$$

Therefore:

$$P=\frac{40}{100} =\boxed{\frac25}$$

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14. Geometric Probability — Area

• Formula:

$$P=\frac{\text{Favorable area}}{\text{Total area}}$$

• Example: A point is selected randomly inside a square of side 10 cm. A circle of radius 5 cm is inscribed in the square. Find the probability that the point lies inside the circle. Solution: Square area:

$$10^2=100$$

Circle area:

$$\pi(5)^2=25\pi$$

Therefore:

$$P=\frac{25\pi}{100} =\boxed{\frac{\pi}{4}}$$

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15. Probability Using Counting

• Formula:

$$P(E)=\frac{\text{Favorable arrangements/selections}} {\text{Total arrangements/selections}}$$

Use combinations when order does not matter.

• Example: Two people are selected from 5 men and 3 women. Find the probability that both selected are women. Solution: Total ways:

$${}^8C_2=28$$

Favorable ways:

$${}^3C_2=3$$

Therefore:

$$P=\frac3{28} =\boxed{\frac3{28}}$$

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16. Probability of Complementary Conditions

• Formula:

$$P(A')=1-P(A)$$

This is especially useful for conditions such as “at least”, “not all”, and “none”.

• Example: Three dice are rolled. Find the probability that not all three show 6. Solution: Probability that all three show 6:

$$\left(\frac16\right)^3=\frac1{216}$$

Therefore:

$$P(\text{not all 6}) =1-\frac1{216} =\boxed{\frac{215}{216}}$$

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Advanced Variants

17. Conditional Probability with Multiple Draws

• Formula:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

For sequential draws, calculate each conditional probability after updating the contents.

• Example: A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Given that the first ball is red, find the probability that both balls are red. Solution: Given the first ball is red, 3 red balls and 6 blue balls remain.

Total remaining:

$$9$$

Therefore:

$$P(\text{second red}\mid\text{first red}) =\boxed{\frac13}$$

Since the first red is already known, the required conditional probability is (\boxed{\frac13}).

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18. At Least One Success in Repeated Trials

• Formula:

$$P(\text{at least one})=1-(1-p)^n$$

• Example: A machine produces a defective item with probability (0.02). Find the probability that at least one defective item occurs among 5 independently produced items. Solution: Probability of a non-defective item:

$$1-0.02=0.98$$

Probability that all 5 are non-defective:

$$(0.98)^5$$

Therefore:

$$P(\text{at least one defective}) =1-(0.98)^5$$ $$\approx\boxed{0.0961}$$

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19. Exactly One Success

• Formula: For (n) independent trials with success probability (p):

$$P(\text{exactly one})= {}^nC_1p(1-p)^{n-1}$$

• Example: A coin is tossed 5 times. Find the probability of getting exactly one head. Solution:

$$P= {}^5C_1 \left(\frac12\right) \left(\frac12\right)^4$$ $$=5\times\frac1{32} =\boxed{\frac5{32}}$$

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20. Probability of At Most / At Least (r) Successes

• Formula:

$$P(X\le r)=\sum_{k=0}^{r}{}^nC_kp^kq^{n-k}$$ $$P(X\ge r)=1-P(X<r)$$

• Example: A coin is tossed 4 times. Find the probability of getting at least 3 heads. Solution:

$$P(X\ge3)=P(X=3)+P(X=4)$$ $$={}^{4}C_3\left(\frac12\right)^4+ {}^{4}C_4\left(\frac12\right)^4$$ $$=\frac4{16}+\frac1{16} =\boxed{\frac5{16}}$$

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21. Probability of Drawing Cards of a Particular Type

• Formula: Without replacement:

$$P(\text{required sequence}) =\frac{\text{favorable sequential outcomes}} {\text{total sequential outcomes}}$$

Or use combinations when only the final selection matters:

$$P=\frac{{}^{\text{favorable}}C_r}{{}^{\text{total}}C_r}$$

• Example: Two cards are drawn from a standard deck without replacement. Find the probability that both are aces. Solution: There are 4 aces among 52 cards:

$$P=\frac4{52}\times\frac3{51}$$ $$=\frac{12}{2652} =\boxed{\frac1{221}}$$

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22. Exactly One of Two Events

• Formula:

$$P(\text{exactly one of }A,B) =P(A)+P(B)-2P(A\cap B)$$

If A and B are independent:

$$=P(A)(1-P(B))+P(B)(1-P(A))$$

• Example: Two dice are rolled. Let A be the event that the first die shows an even number and B that the second die shows an even number. Find the probability that exactly one die is even. Solution:

$$P(A)=P(B)=\frac12$$

Since the dice are independent:

$$P(\text{exactly one}) =\frac12\cdot\frac12+\frac12\cdot\frac12$$ $$=\boxed{\frac12}$$

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23. Odds in Favor and Odds Against

• Formula: If

$$P(E)=\frac ab,$$

then:

$$\text{Odds in favor}=a:(b-a)$$ $$\text{Odds against}=(b-a):a$$

• Example: The probability of winning a game is (\frac35). Find the odds in favor of winning. Solution: Probability of winning:

$$\frac35$$

Probability of losing:

$$1-\frac35=\frac25$$

Therefore:

$$\text{Odds in favor}=3:2$$

Answer = (\boxed{3:2}).

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24. Probability from Odds

• Formula: If odds in favor of an event are (a:b):

$$P(E)=\frac{a}{a+b}$$

If odds against are (a:b):

$$P(E)=\frac{b}{a+b}$$

• Example: The odds against a candidate winning an election are (3:2). Find the probability that the candidate wins. Solution: Odds against:

$$3:2$$

Therefore favorable part:

$$2$$

Total:

$$3+2=5$$

Hence:

$$P(\text{win})=\boxed{\frac25}$$

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25. Bayes’ Theorem with Defective Products

• Formula:

$$P(\text{source}\mid\text{defective}) ==================================== \frac{P(\text{source})P(\text{defective}\mid\text{source})} {P(\text{defective})}$$

• Example: Factory A produces 60% of items with a 2% defect rate. Factory B produces 40% with a 5% defect rate. An item is found defective. Find the probability it came from B. Solution:

$$P(B)=0.4,\qquad P(D\mid B)=0.05$$

Total defective probability:

$$P(D)=0.6(0.02)+0.4(0.05)$$ $$=0.012+0.020=0.032$$

Therefore:

$$P(B\mid D) =\frac{0.4(0.05)}{0.032}$$ $$=\frac{0.020}{0.032} =\boxed{\frac58}$$

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26. Geometric Probability — Distance from a Point

• Formula: For a uniformly selected point on a line segment, probability is the ratio of favorable length to total length.
• Example: A point is chosen randomly on a 10 cm line segment. Find the probability that its distance from the left endpoint is less than 3 cm. Solution: Favorable length:

$$3\text{ cm}$$

Total length:

$$10\text{ cm}$$

Therefore:

$$P=\frac3{10} =\boxed{\frac3{10}}$$