Step 1: Theoretical Analysis of the Subset Condition

Given the condition $A \subset B$, by the definition of a subset in probability theory, every sample point in event $A$ is also contained within event $B$. Consequently, the intersection of $A$ and $B$ is simply $A$:

$A \cap B = A$ [Definition of Subset]

Therefore, the probability of the intersection is:

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

Step 2: Evaluating Conditional Probability

The conditional probability of $A$ given $B$ is defined by the formula [Per the Axioms of Probability]:

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

Substituting the result from Step 1 into this formula:

$P(A|B) = \frac{P(A)}{P(B)}$

Step 3: Logical Deduction of the Inequality

Since $A \subset B$, it follows that $P(A) \leq P(B)$ [Monotonicity Property of Probability]. Given $P(B) \neq 0$, we analyze the possible values for $P(A|B)$:

• Since $P(A) \leq P(B)$, then $\frac{P(A)}{P(B)} \leq 1$.
• Since $P(A) \geq 0$ and $P(B) > 0$, then $\frac{P(A)}{P(B)} \geq 0$.

Step 4: Comparative Analysis of Options

In standard textbook problems of this type, the objective is to identify the relationship between $P(A|B)$ and $P(A)$.

Final Solution: The correct relationship is $P(A|B) \geq P(A)$, as $P(A|B) = \frac{P(A)}{P(B)}$ and $P(B) \leq 1$.

Step 1: Defining Events and Initial Probabilities

Let the events be defined as follows:

• $E_1$: The event of selecting Bag I.
• $E_2$: The event of selecting Bag II.
• $A$: The event of drawing a red ball.

Since the bag is selected at random, the prior probabilities are:

• $P(E_1) = \frac{1}{2}$ [Per the definition of random selection]
• $P(E_2) = \frac{1}{2}$ [Per the definition of random selection]

Step 2: Calculating Conditional Probabilities

Based on the contents of the bags:

• Bag I contains $4$ red and $4$ black balls (Total = $8$ balls).
• Bag II contains $2$ red and $6$ black balls (Total = $8$ balls).

The conditional probabilities of drawing a red ball given the selected bag are:

• $P(A|E_1) = \frac{4}{8} = \frac{1}{2}$ [By the classical definition of probability]
• $P(A|E_2) = \frac{2}{8} = \frac{1}{4}$ [By the classical definition of probability]

Step 3: Calculating Total Probability of Drawing a Red Ball

Using the Law of Total Probability [Per the Law of Total Probability]:

Step 4: Applying Bayes' Theorem

We seek the posterior probability $P(E_1|A)$, the probability that the ball came from Bag I given it is red [Per Bayes' Theorem]:

Visual Representation of Probability Tree

Final Solution: The probability that the ball is drawn from the first bag is $\frac{2}{3}$.

Step 1: Define Events and Initial Probabilities

Let $E_1$ be the event that the first ball drawn is Red, and $E_2$ be the event that the first ball drawn is Black. The urn initially contains 5 Red ($R$) and 5 Black ($B$) balls, totaling 10 balls.

• $P(E_1) = \frac{5}{10} = \frac{1}{2}$
• $P(E_2) = \frac{5}{10} = \frac{1}{2}$

Step 2: Define Conditional Probabilities for the Second Draw

Let $A$ be the event that the second ball drawn is Red. We must consider the state of the urn after the first ball is replaced and two additional balls of the same color are added.

Step 3: Apply the Law of Total Probability

To find the total probability of event $A$, we use the Law of Total Probability: $P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

Substituting the values derived above:

$P(A) = \left( \frac{1}{2} \times \frac{7}{12} \right) + \left( \frac{1}{2} \times \frac{5}{12} \right)$

Step 4: Algebraic Simplification

$P(A) = \frac{7}{24} + \frac{5}{24}$

$P(A) = \frac{12}{24}$

$P(A) = \frac{1}{2}$

Final Solution: The probability that the second ball is red is $\frac{1}{2}$.

Step 1: Defining Events and Given Probabilities

Let the sample space be the set of all students in the college. We define the following events:

• $H$: The event that the chosen student resides in a hostel.
• $D$: The event that the chosen student is a day scholar.
• $A$: The event that the chosen student attains an A grade.

Based on the provided data, we assign the following probabilities:

• $P(H) = 0.60$ [Given: 60% of students are hostellers]
• $P(D) = 0.40$ [Given: 40% of students are day scholars]
• $P(A|H) = 0.30$ [Given: 30% of hostellers attain A grade]
• $P(A|D) = 0.20$ [Given: 20% of day scholars attain A grade]

Step 2: Visualization of Probability Space

Step 3: Calculating Total Probability of Event A

To find the probability that a randomly chosen student has an A grade, we use the Law of Total Probability [By Theorem of Total Probability]:

$P(A) = P(H) \cdot P(A|H) + P(D) \cdot P(A|D)$

$P(A) = (0.60 \times 0.30) + (0.40 \times 0.20)$

$P(A) = 0.18 + 0.08$

$P(A) = 0.26$

Step 4: Applying Bayes' Theorem

We are required to find the probability that the student is a hosteller, given that they have an A grade. This is represented by $P(H|A)$. [By Bayes' Theorem]:

$P(H|A) = \frac{P(H) \cdot P(A|H)}{P(A)}$

$P(H|A) = \frac{0.60 \times 0.30}{0.26}$

$P(H|A) = \frac{0.18}{0.26}$

Step 5: Simplification of the Fraction

$P(H|A) = \frac{18}{26} = \frac{9}{13}$

Converting to a decimal (approximate): $P(H|A) \approx 0.6923$

Final Solution: The probability that the student is a hosteller is $\frac{9}{13}$.

Step 1: Definition of Events

Let the sample space be the set of all items produced by the factory. We define the following events:

• $E_1$: The event that the item is produced by Machine A.
• $E_2$: The event that the item is produced by Machine B.
• $A$: The event that the chosen item is defective.

Step 2: Assigning Probabilities to Events

Based on the provided data, we define the prior probabilities [Per the Law of Total Probability]:

• $P(E_1) = 60\% = 0.60$
• $P(E_2) = 40\% = 0.40$

The conditional probabilities of producing a defective item given the machine are:

• $P(A|E_1) = 2\% = 0.02$
• $P(A|E_2) = 1\% = 0.01$

Step 3: Calculating Total Probability of a Defective Item

The total probability $P(A)$ that an item is defective is given by the formula [By the Law of Total Probability]:

$P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

Substituting the values:

$P(A) = (0.60 \times 0.02) + (0.40 \times 0.01)$

$P(A) = 0.012 + 0.004 = 0.016$

Step 4: Applying Bayes' Theorem

We are required to find the probability that the defective item came from Machine B, which is $P(E_2|A)$. [By Bayes' Theorem]:

$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

$P(E_2|A) = \frac{0.40 \times 0.01}{0.016}$

$P(E_2|A) = \frac{0.004}{0.016}$

$P(E_2|A) = \frac{4}{16} = \frac{1}{4} = 0.25$

Visual Representation of Probability Distribution

Final Solution: The probability that the defective item was produced by machine B is $0.25$ or $25\%$.

Step 1: Defining Events and Sample Spaces

Let $E_1$ be the event that the girl gets 5 or 6 on the die. The probability of this event is $P(E_1) = \frac{2}{6} = \frac{1}{3}$.

Let $E_2$ be the event that the girl gets 1, 2, 3, or 4 on the die. The probability of this event is $P(E_2) = \frac{4}{6} = \frac{2}{3}$.

Let $A$ be the event that she obtains exactly one head.

Step 2: Calculating Conditional Probabilities

If $E_1$ occurs (she gets 5 or 6), she tosses a coin 3 times. The number of heads follows a binomial distribution $B(n=3, p=0.5)$. The probability of getting exactly one head is:

$P(A|E_1) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \times \frac{1}{8} = \frac{3}{8}$

If $E_2$ occurs (she gets 1, 2, 3, or 4), she tosses a coin once. The probability of getting exactly one head is:

$P(A|E_2) = \binom{1}{1} \left(\frac{1}{2}\right)^1 = \frac{1}{2}$

Step 3: Visual Representation of Probability Tree

Step 4: Applying Bayes' Theorem

We are required to find $P(E_2|A)$. According to Bayes' Theorem:

$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

Substituting the values calculated in steps 1 and 2:

$P(E_2|A) = \frac{\frac{2}{3} \cdot \frac{1}{2}}{\left(\frac{1}{3} \cdot \frac{3}{8}\right) + \left(\frac{2}{3} \cdot \frac{1}{2}\right)}$

Step 5: Algebraic Simplification

Numerator: $\frac{2}{6} = \frac{1}{3}$

Denominator: $\left(\frac{3}{24}\right) + \left(\frac{2}{6}\right) = \frac{1}{8} + \frac{1}{3}$

Finding common denominator for the sum: $\frac{3 + 8}{24} = \frac{11}{24}$

$P(E_2|A) = \frac{1/3}{11/24} = \frac{1}{3} \times \frac{24}{11} = \frac{8}{11}$

Final Solution: The probability that she threw 1, 2, 3, or 4 given that she obtained exactly one head is $\frac{8}{11}$.

Step 1: Definition of Events and Initial Probabilities

Let the events be defined as follows:

• $E_1$: The first group wins the election.
• $E_2$: The second group wins the election.
• $A$: A new product is introduced.

Based on the provided data, the prior probabilities are:

• $P(E_1) = 0.6$
• $P(E_2) = 0.4$

[Justification: The sum of probabilities for mutually exclusive and exhaustive events must equal 1, i.e., $0.6 + 0.4 = 1$.]

Step 2: Conditional Probabilities

The probabilities of introducing a new product given the winning group are:

• $P(A|E_1) = 0.7$ (Probability of new product given group 1 wins)
• $P(A|E_2) = 0.3$ (Probability of new product given group 2 wins)

Step 3: Calculating Total Probability of Event A

By the Law of Total Probability, the probability that a new product is introduced is calculated as:

$P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

$P(A) = (0.6 \times 0.7) + (0.4 \times 0.3)$

$P(A) = 0.42 + 0.12 = 0.54$

Step 4: Applying Bayes' Theorem

We need to find the probability that the new product was introduced by the second group, given that a new product was introduced. We seek $P(E_2|A)$.

[Per Bayes' Theorem: $P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$]

$P(E_2|A) = \frac{0.4 \times 0.3}{0.54}$

$P(E_2|A) = \frac{0.12}{0.54}$

Step 5: Simplification

$P(E_2|A) = \frac{12}{54} = \frac{2}{9}$

Final Solution: The probability that the new product was introduced by the second group is $\frac{2}{9}$ or approximately $0.222$.

Step 1: Definition of Events

Let $E$ be the event that the item produced is defective. Let $A_1, A_2,$ and $A_3$ be the events that the items are produced by operators A, B, and C respectively.

• $P(A_1) = 50\% = 0.5$
• $P(A_2) = 30\% = 0.3$
• $P(A_3) = 20\% = 0.2$

[By the Principle of Exhaustive Events, $\sum P(A_i) = 0.5 + 0.3 + 0.2 = 1.0$]

Step 2: Conditional Probabilities of Defectives

The probability of producing a defective item given the operator is as follows:

• $P(E|A_1) = 1\% = 0.01$
• $P(E|A_2) = 5\% = 0.05$
• $P(E|A_3) = 7\% = 0.07$

Step 3: Calculation of Total Probability $P(E)$

Using the Law of Total Probability, the total probability that a randomly chosen item is defective is given by: [Per the Law of Total Probability]

Step 4: Applying Bayes' Theorem

To find the probability that the defective item was produced by operator A, we calculate $P(A_1|E)$ using Bayes' Theorem: [Per Bayes' Theorem]

Final Solution: The probability that the defective item was produced by operator A is $\frac{5}{34}$ or approximately $0.147$.

Step 1: Definition of Events

Let the sample space be the set of all persons tested. We define the following events:

• $E_1$: The person has the disease.
• $E_2$: The person is healthy (does not have the disease).
• $A$: The test result is positive.

Step 2: Assigning Probabilities

Based on the data provided:

• $P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001$
• $P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0.999$
• $P(A|E_1)$: Probability of a positive test given the disease is present = $99\% = 0.99$
• $P(A|E_2)$: Probability of a positive test given the person is healthy (false positive) = $0.5\% = 0.005$

Step 3: Applying Bayes' Theorem

We aim to find $P(E_1|A)$, the probability that the person has the disease given a positive test result. By Bayes' Theorem [per the Law of Total Probability and Conditional Probability axioms]:

$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$

Step 4: Calculation

Substitute the assigned values into the equation:

Therefore:

$P(E_1|A) = \frac{0.00099}{0.005985}$

$P(E_1|A) = \frac{990}{5985}$

Step 5: Simplification

Dividing both terms by their greatest common divisor ($45$):

$P(E_1|A) = \frac{990 \div 45}{5985 \div 45} = \frac{22}{133}$

Final Solution: The probability that a person has the disease given that their test result is positive is $\frac{22}{133}$ or approximately $0.1654$.

Step 1: Defining Events and Given Probabilities

Let the events be defined as follows:

• $E_1$: The student knows the answer.
• $E_2$: The student guesses the answer.
• $A$: The student answers the question correctly.

Based on the problem statement, we have the following probabilities:

• $P(E_1) = \frac{3}{4}$
• $P(E_2) = \frac{1}{4}$
• $P(A|E_1) = 1$ (If the student knows the answer, the probability of being correct is $1$ [Certainty Axiom])
• $P(A|E_2) = \frac{1}{4}$ (Given probability of guessing correctly)

Step 2: Calculating Total Probability of a Correct Answer

To find the total probability that the student answers correctly, $P(A)$, we use the Law of Total Probability [Law of Total Probability]:

$P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

Substituting the values:

$P(A) = \left( \frac{3}{4} \times 1 \right) + \left( \frac{1}{4} \times \frac{1}{4} \right)$

$P(A) = \frac{3}{4} + \frac{1}{16}$

$P(A) = \frac{12}{16} + \frac{1}{16} = \frac{13}{16}$

Step 3: Applying Bayes' Theorem

We are required to find the probability that the student knows the answer given that they answered correctly, denoted as $P(E_1|A)$. Per Bayes' Theorem [Bayes' Theorem]:

$P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{P(A)}$

Substitute the established values into the formula:

$P(E_1|A) = \frac{\frac{3}{4} \times 1}{\frac{13}{16}}$

$P(E_1|A) = \frac{3}{4} \times \frac{16}{13}$

$P(E_1|A) = \frac{3 \times 4}{13} = \frac{12}{13}$

Visual Representation of Probability Tree

Final Solution: The probability that the student knows the answer given that they answered correctly is $\frac{12}{13}$.

Step 1: Identification of Events and Total Population

Let the total number of insured persons be $N$. Given:

• Number of scooter drivers ($n_1$) = $2000$
• Number of car drivers ($n_2$) = $4000$
• Number of truck drivers ($n_3$) = $6000$
• Total drivers ($N$) = $2000 + 4000 + 6000 = 12000$

Define the events:

• $E_1$: The selected person is a scooter driver.
• $E_2$: The selected person is a car driver.
• $E_3$: The selected person is a truck driver.
• $A$: The selected person meets with an accident.

Step 2: Assigning Probabilities [By Classical Definition of Probability]

The probability of selecting a driver from each category is calculated as $P(E_i) = \frac{n_i}{N}$:

• $P(E_1) = \frac{2000}{12000} = \frac{1}{6}$
• $P(E_2) = \frac{4000}{12000} = \frac{2}{6} = \frac{1}{3}$
• $P(E_3) = \frac{6000}{12000} = \frac{3}{6} = \frac{1}{2}$

Step 3: Conditional Probabilities of Accidents

The conditional probabilities of an accident given the type of driver are provided as:

• $P(A|E_1) = 0.01 = \frac{1}{100}$
• $P(A|E_2) = 0.03 = \frac{3}{100}$
• $P(A|E_3) = 0.15 = \frac{15}{100}$

Step 4: Applying Bayes' Theorem

We seek the posterior probability $P(E_1|A)$, which is defined by Bayes' Theorem as:

$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{\sum_{i=1}^{3} P(E_i)P(A|E_i)}$

Step 5: Calculation of the Total Probability $P(A)$

$P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)$

$P(A) = \left(\frac{1}{6} \times \frac{1}{100}\right) + \left(\frac{2}{6} \times \frac{3}{100}\right) + \left(\frac{3}{6} \times \frac{15}{100}\right)$

$P(A) = \frac{1 + 6 + 45}{600} = \frac{52}{600}$

Step 6: Final Posterior Probability Calculation

$P(E_1|A) = \frac{\frac{1}{600}}{\frac{52}{600}} = \frac{1}{52}$

Final Solution: The probability that the person is a scooter driver given that they met with an accident is $\frac{1}{52}$.

Step 1: Defining Events and Given Probabilities

Let the events be defined as follows:

• Let $H$ be the event that a Head appears on the coin. $P(H) = \frac{1}{2}$
• Let $T$ be the event that a Tail appears on the coin. $P(T) = \frac{1}{2}$
• Let $E$ be the event that $A$ reports that a head appears.

The conditional probabilities regarding the truthfulness of $A$ are:

• Probability that $A$ speaks truth is $P(\text{Truth}) = \frac{4}{5}$
• Probability that $A$ lies is $P(\text{Lie}) = 1 - \frac{4}{5} = \frac{1}{5}$

Step 2: Calculating Conditional Likelihoods

Using the definition of truthfulness, we determine $P(E|H)$ and $P(E|T)$:

• $P(E|H)$ (A reports head when head actually appeared) = $P(\text{Truth}) = \frac{4}{5}$
• $P(E|T)$ (A reports head when tail actually appeared) = $P(\text{Lie}) = \frac{1}{5}$

Step 3: Applying Bayes' Theorem

We need to find $P(H|E)$, the probability that a Head appeared given that $A$ reported a Head. Per Bayes' Theorem [Theorem of Total Probability and Posterior Probability]:

$P(H|E) = \frac{P(H) \cdot P(E|H)}{P(H) \cdot P(E|H) + P(T) \cdot P(E|T)}$

Step 4: Substituting Values and Algebraic Simplification

Substitute the established probabilities into the theorem:

$P(H|E) = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{4}{5}\right)}{\left(\frac{1}{2}\right) \cdot \left(\frac{4}{5}\right) + \left(\frac{1}{2}\right) \cdot \left(\frac{1}{5}\right)}$

Factor out $\frac{1}{2}$ from the denominator:

$P(H|E) = \frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2} \cdot \left(\frac{4}{5} + \frac{1}{5}\right)}$

$P(H|E) = \frac{\frac{4}{5}}{\frac{5}{5}} = \frac{4}{5}$

Visual Representation of the Probability Space

Final Solution: The probability that actually there was a head is $\frac{4}{5}$.

Step 1: Defining Events and Prior Probabilities

Let $C_1, C_2,$ and $C_3$ be the events of choosing the two-headed coin, the biased coin, and the unbiased coin, respectively. Since one coin is chosen at random from three:

• $P(C_1) = \frac{1}{3}$
• $P(C_2) = \frac{1}{3}$
• $P(C_3) = \frac{1}{3}$

[Per the definition of equally likely outcomes for selecting one out of three options].

Step 2: Determining Conditional Probabilities of Heads (Event H)

Let $H$ be the event that the chosen coin shows heads. We determine the probability of $H$ given each coin choice:

• For the two-headed coin ($C_1$): $P(H|C_1) = 1$ [Since both sides are heads].
• For the biased coin ($C_2$): $P(H|C_2) = \frac{75}{100} = \frac{3}{4}$ [Given].
• For the unbiased coin ($C_3$): $P(H|C_3) = \frac{1}{2}$ [Standard probability for a fair coin].

Step 3: Applying the Total Probability Theorem

To find the overall probability of getting heads, $P(H)$, we sum the joint probabilities:

$P(H) = P(C_1)P(H|C_1) + P(C_2)P(H|C_2) + P(C_3)P(H|C_3)$

$P(H) = \left(\frac{1}{3} \times 1\right) + \left(\frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right)$

$P(H) = \frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4+3+2}{12} = \frac{9}{12} = \frac{3}{4}$

Step 4: Applying Bayes' Theorem

We seek the posterior probability $P(C_1|H)$, the probability the coin was two-headed given that a head appeared.

$P(C_1|H) = \frac{P(C_1)P(H|C_1)}{P(H)}$

$P(C_1|H) = \frac{\frac{1}{3} \times 1}{\frac{3}{4}}$

$P(C_1|H) = \frac{1}{3} \times \frac{4}{3} = \frac{4}{9}$

[Per Bayes' Theorem, which relates conditional probabilities and prior knowledge].

Final Solution: The probability that the coin was the two-headed coin is $\frac{4}{9}$.

Step 1: Defining Events and Prior Probabilities

Let the sample space consist of $52$ cards. We define the following events based on the lost card:

• $E_1$: The lost card is a diamond.
• $E_2$: The lost card is not a diamond.
• $A$: The event that two cards drawn from the remaining $51$ cards are both diamonds.

Given the composition of a standard deck ($13$ diamonds, $39$ non-diamonds):

• $P(E_1) = \frac{13}{52} = \frac{1}{4}$
• $P(E_2) = \frac{39}{52} = \frac{3}{4}$

Step 2: Determining Conditional Probabilities

We calculate the probability of drawing two diamonds ($A$) given the status of the lost card:

• If $E_1$ occurs: The remaining deck has $51$ cards, containing $12$ diamonds. The probability of picking two diamonds is: $P(A|E_1) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}$
• If $E_2$ occurs: The remaining deck has $51$ cards, containing $13$ diamonds. The probability of picking two diamonds is: $P(A|E_2) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}$

Step 3: Applying Bayes' Theorem

We seek $P(E_1|A)$, the probability the lost card was a diamond given that two diamonds were drawn [Per Bayes' Theorem]:

$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$

Step 4: Algebraic Substitution and Calculation

Substitute the values calculated in Step 1 and 2:

$P(E_1|A) = \frac{(\frac{1}{4}) \times (\frac{132}{2550})}{(\frac{1}{4}) \times (\frac{132}{2550}) + (\frac{3}{4}) \times (\frac{156}{2550})}$

Since the denominator $(\frac{1}{4} \times 2550)$ is common, we simplify:

$P(E_1|A) = \frac{132}{132 + (3 \times 156)}$

$P(E_1|A) = \frac{132}{132 + 468} = \frac{132}{600}$

Step 5: Final Simplification

Reducing the fraction by dividing by the greatest common divisor ($12$):

$\frac{132 \div 12}{600 \div 12} = \frac{11}{50}$

Final Solution: The probability that the lost card was a diamond is $\frac{11}{50}$ or $0.22$.