GSEB Class 12 Maths Solutions Chapter 13 સંભાવના Exercise 13.4

Gseb Solutions Class 12 Maths Chapter 13 संभावना Ex 13.4

Gujarat Board Textbook Solutions Class 12 Maths Chapter 13 संभावना Ex 13.4

Question 1. નીચેના પૈકી કર્યાં વિતરણ યાદૈચ્છિક ચલનાં સંભાવના વિતરણ નથી તે લખો. તમારા જવાબનું સમર્થન કરો :
(i)

X	0	1	2
P(X)	0.4	0.4	0.2

(ii)

X	0	1	2	3	4
P(X)	0.1	0.5	0.2	-0.1	0.3

(iii)

Y	-1	0	1
P(Y)	0.6	0.1	0.2

(iv)

Z	3	2	1	0	-1
P(Z)	0.3	0.2	0.4	0.1	0.05

Answer:
(i) For a probability distribution, all probabilities must be non-negative, and their sum must be 1. Here, \(P(X = 0) + P(X = 1) + P(X = 2) = 0.4 + 0.4 + 0.2 = 1\). Also, all \(P(x_i) \geq 0\) for \(i = 0, 1, 2\). Therefore, the given distribution is a probability distribution for the random variable X.
(ii) For a probability distribution, all probabilities must be non-negative. Here, \(P(X = 3) = -0.1\), which is less than 0. Probabilities cannot be negative. Therefore, the given distribution is not a probability distribution for the random variable X.
(iii) For a probability distribution, the sum of all probabilities must be 1. Here, \(P(Y = -1) + P(Y = 0) + P(Y = 1) = 0.6 + 0.1 + 0.2 = 0.9\). Since the sum 0.9 is not equal to 1, this distribution is not a probability distribution for the random variable Y.
(iv) For a probability distribution, the sum of all probabilities must be 1. Here, \(P(Z = 3) + P(Z = 2) + P(Z = 1) + P(Z = 0) + P(Z = -1) = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05\). Since the sum 1.05 is not equal to 1, this distribution is not a probability distribution for the random variable Z.In simple words: A probability distribution must have all probabilities as positive numbers or zero, and all probabilities must add up to exactly one. We check these two rules for each table to see if it's a valid distribution. If any rule is broken, it's not a probability distribution.

🎯 Exam Tip: When identifying a probability distribution, always verify two conditions: (1) each probability value P(X=x_i) must be non-negative, and (2) the sum of all probabilities ΣP(X=x_i) must equal 1. Failing either condition means it's not a valid distribution.

 

Question 2. એક પાત્રમાં 5 લાલ રંગના અને 2 કાળા રંગના દડા છે. બે દડા યાદૈચ્છિક રીતે પસંદ કરવામાં આવે છે. જો X કાળા રંગના દડાઓની સંખ્યા દર્શાવે છે. X ની શક્ય કિંમતો કઈ-કઈ છે ? શું X યાદૈચ્છિક ચલ છે ?
Answer: There are 5 red balls and 2 black balls in a container. Two balls are picked randomly. Let R stand for a red ball and B for a black ball. The ways to pick two balls are RR, RB, BR, and BB.
X shows the number of black balls picked.
- If X = 0, it means no black balls are picked (RR).
- If X = 1, it means one black ball is picked (RB or BR).
- If X = 2, it means two black balls are picked (BB).
So, the possible values for X are 0, 1, and 2.
Yes, X is a random variable because its values depend on the random outcome of picking balls.In simple words: We have red and black balls. We pick two. X tells us how many black balls we got. X can be 0, 1, or 2 black balls. Since we pick randomly, X is a random variable.

🎯 Exam Tip: To find the possible values of a random variable, consider all possible outcomes of the experiment and identify what the random variable represents for each outcome. A random variable assigns a numerical value to each outcome in a sample space.

 

Question 3. ધારો કે જ્યારે સિક્કાને 6 વખત ઉછાળવામાં આવે છે ત્યારે X એ છાપની સંખ્યા અને કાંટાની સંખ્યાનો તફાવત દર્શાવે છે. X ની શક્ય કિંમતો શું છે ?
Answer: Suppose a coin is tossed 6 times. We want to find the possible values of X, which is the difference between the number of heads and the number of tails.
The total number of tosses is 6. Let H be the number of heads and T be the number of tails.
We know that \(H + T = 6\).
The variable X is defined as \(X = |H - T|\).
Let's list the possible combinations for H and T:

Heads (H)	6	5	4	3	2	1	0
Tails (T)	0	1	2	3	4	5	6
Difference \(|H - T|\)	6	4	2	0	2	4	6

From the table, the possible values for X (the difference between heads and tails) are 0, 2, 4, and 6.In simple words: We toss a coin 6 times. X is the difference between how many heads and how many tails we get. We list all ways to get heads and tails (like 6 heads and 0 tails, or 3 heads and 3 tails) and then find the difference for each. The possible differences are 0, 2, 4, and 6.

🎯 Exam Tip: For problems involving coin tosses, systematically list the number of heads and tails. The sum of heads and tails will always equal the total number of tosses. Remember that the difference is usually an absolute value, meaning it's always positive or zero.

 

Question 4. નીચેના કિસ્સાઓમાં સંભાવના વિતરણ શોધો.
(i) સિક્કાને બે વખત ઉછાળતાં મળતી છાપની સંખ્યા
(ii) ત્રણ સિક્કાઓને એકસાથે ઉછાળતાં મળતી કાંટાની સંખ્યા
(iii) સિક્કાને ચાર વખત ઉછાળતાં મળતી છાપની સંખ્યા
Answer:
(i) We toss a coin two times.
The sample space (all possible outcomes) is \(S = \{HH, HT, TH, TT\}\).
Let X be the number of heads. The possible values for X are 0, 1, and 2.
- For \(X = 0\) (no heads): This occurs with outcome TT. So, \(P(X = 0) = P(TT) = \frac{1}{4}\).
- For \(X = 1\) (one head): This occurs with outcomes HT or TH. So, \(P(X = 1) = P(HT \text{ અથવા } TH) = \frac{2}{4} = \frac{1}{2}\).
- For \(X = 2\) (two heads): This occurs with outcome HH. So, \(P(X = 2) = P(HH) = \frac{1}{4}\).
The probability distribution of the random variable X is:

X	0	1	2
P(X)	\(\frac{1}{4}\)	\(\frac{1}{2}\)	\(\frac{1}{4}\)

(ii) We toss three coins at the same time.
The sample space is \(S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}\).
Let X be the number of tails. The possible values for X are 0, 1, 2, and 3.
- For \(X = 0\) (no tails): This occurs with outcome HHH. So, \(P(X = 0) = P(HHH) = \frac{1}{8}\).
- For \(X = 1\) (one tail): This occurs with outcomes HHT, HTH, or THH. So, \(P(X = 1) = P(HHT \text{ અથવા } HTH \text{ અથવા } THH) = \frac{3}{8}\).
- For \(X = 2\) (two tails): This occurs with outcomes HTT, THT, or TTH. So, \(P(X = 2) = P(HTT \text{ અથવા } THT \text{ અથવા } TTH) = \frac{3}{8}\).
- For \(X = 3\) (three tails): This occurs with outcome TTT. So, \(P(X = 3) = P(TTT) = \frac{1}{8}\).
The probability distribution of the random variable X is:

X	0	1	2	3
P(X)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)

(iii) We toss a coin four times.
The total number of possible outcomes is \(n(S) = 2^4 = 16\).
Let X be the number of heads on the coin. The possible values for X are 0, 1, 2, 3, and 4.
- For \(X = 0\) (no heads): This occurs with outcome TTTT. So, \(P(X = 0) = P(TTTT) = \frac{1}{16}\).
- For \(X = 1\) (one head): This occurs with outcomes HTTT, THTT, TTHT, TTTH. So, \(P(X = 1) = \frac{4}{16} = \frac{1}{4}\).
- For \(X = 2\) (two heads): This occurs with outcomes HHTT, HTH T, HTTH, THHT, THTH, TTHH. So, \(P(X = 2) = \frac{6}{16} = \frac{3}{8}\).
- For \(X = 3\) (three heads): This occurs with outcomes HHHT, HHTH, HTHH, THHH. So, \(P(X = 3) = \frac{4}{16} = \frac{1}{4}\).
- For \(X = 4\) (four heads): This occurs with outcome HHHH. So, \(P(X = 4) = P(HHHH) = \frac{1}{16}\).
The probability distribution of the random variable X is:

X	0	1	2	3	4
P(X)	\(\frac{1}{16}\)	\(\frac{1}{4}\)	\(\frac{3}{8}\)	\(\frac{1}{4}\)	\(\frac{1}{16}\)

In simple words: We find all possible outcomes for tossing coins a certain number of times. Then, we count how many times the specific event (like getting a head or a tail) happens. We divide that count by the total outcomes to get the probability for each value of X. Finally, we put these into a table.

🎯 Exam Tip: When dealing with coin toss problems, remember that each toss is an independent event. The sample space grows as \(2^n\) (where n is the number of tosses). Be careful to list all outcomes correctly to calculate probabilities accurately, especially for the number of heads or tails.

 

Question 5. જો સફળતા (i) 4 કરતાં મોટી સંખ્યા (ii) ઓછામાં ઓછા એક પાસા પર પૂર્ણાંક 6 મળે, એ પ્રમાણે વ્યાખ્યાયિત થતી હોય તો પાસાને બે વખત ઉછાળવામાં સફળતા માટેનું સંભાવના વિતરણ શોધો.
Answer:
(i) Success is defined as getting a number greater than 4 on a single die. The pair of dice is thrown twice.
When a single die is rolled, the sample space is \(S = \{1, 2, 3, 4, 5, 6\}\), so \(n(S) = 6\).
Let A be the event of getting a number greater than 4, i.e., \(A = \{5, 6\}\).
The probability of success is \(P(A) = \frac{2}{6} = \frac{1}{3}\).
The probability of failure (not getting a number greater than 4) is \(P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}\).
Now, a pair of dice is thrown twice. Let X be the number of times a "success" (getting a number greater than 4 on *either* die, if we interpret it as total successes in two trials of a single die) occurs. Based on the calculation structure, it considers two independent dice throws. Let X be the number of dice that show a number greater than 4 when *two dice are thrown once*. So the possible values for X are 0, 1, 2.
- For \(X = 0\): Both dice show a number not greater than 4 (failure on both).
\(P(X = 0) = P(A') P(A') = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}\).
- For \(X = 1\): One die shows a number greater than 4, and the other does not.
\(P(X = 1) = P(A)P(A') + P(A')P(A) = \frac{1}{3} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}\).
- For \(X = 2\): Both dice show a number greater than 4 (success on both).
\(P(X = 2) = P(A)P(A) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\).
The probability distribution for X is:

X	0	1	2
P(X)	\(\frac{4}{9}\)	\(\frac{4}{9}\)	\(\frac{1}{9}\)

(ii) Success is defined as getting a 6 on at least one die. We are throwing two dice.
Let A be the event of getting a 6 on a single die. Then \(P(A) = \frac{1}{6}\).
Let A' be the event of not getting a 6 on a single die. Then \(P(A') = 1 - \frac{1}{6} = \frac{5}{6}\).
Let X be the number of 6s obtained when two dice are thrown. The possible values for X are 0, 1, and 2.
- For \(X = 0\): No 6s are obtained on either die (both show non-6).
\(P(X = 0) = P(A')P(A') = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\).
- For \(X = 1\): Exactly one 6 is obtained (one die shows 6, the other does not).
\(P(X = 1) = P(A)P(A') + P(A')P(A) = \frac{1}{6} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{6} = \frac{5}{36} + \frac{5}{36} = \frac{10}{36}\).
- For \(X = 2\): Two 6s are obtained (both dice show 6).
\(P(X = 2) = P(A)P(A) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
The probability distribution for X is:

X	0	1	2
P(X)	\(\frac{25}{36}\)	\(\frac{10}{36}\)	\(\frac{1}{36}\)

In simple words: For each part, we first define what "success" means on one die and find its probability. Then, for throwing two dice, we find the chance of getting zero, one, or two successes. We list these chances in a table.

🎯 Exam Tip: When dealing with multiple independent trials (like rolling two dice), treat each trial's outcome as independent. Remember to account for all possible combinations that lead to a specific number of successes (e.g., success-failure and failure-success for one success).

 

Question 6. 30 વીજળીના ગોળાઓમાંથી 6 ગોળા ખામીયુક્ત છે. પુરવણી સહિત 4 ગોળાઓનો નિદર્શ યાર્દચ્છિક રીતે લીધો છે. ખામીયુક્ત ગોળાઓની સંખ્યા માટેનું સંભાવના વિતરણ શોધી.
Answer: There are 30 electric bulbs in total. 6 bulbs are defective, and \(30 - 6 = 24\) bulbs are non-defective.
A sample of 4 bulbs is drawn randomly with replacement.
Let D be the event that a bulb is defective.
The probability of drawing a defective bulb is \(P(D) = \frac{6}{30} = \frac{1}{5}\).
The probability of drawing a non-defective bulb is \(P(D') = 1 - P(D) = 1 - \frac{1}{5} = \frac{4}{5}\).
Let X be the number of defective bulbs in the sample of 4. Since bulbs are drawn with replacement, this is a binomial distribution. The possible values for X are 0, 1, 2, 3, and 4.
- For \(X = 0\) (no defective bulbs): All 4 bulbs are non-defective.
\(P(X = 0) = (P(D'))^4 = \left(\frac{4}{5}\right)^4 = \frac{256}{625}\).
- For \(X = 1\) (one defective bulb): One bulb is defective, and three are non-defective. There are \(\binom{4}{1}\) ways this can happen.
\(P(X = 1) = \binom{4}{1} P(D)^1 P(D')^3 = 4 \times \frac{1}{5} \times \left(\frac{4}{5}\right)^3 = 4 \times \frac{1}{5} \times \frac{64}{125} = \frac{256}{625}\).
- For \(X = 2\) (two defective bulbs): Two bulbs are defective, and two are non-defective. There are \(\binom{4}{2}\) ways this can happen.
\(P(X = 2) = \binom{4}{2} P(D)^2 P(D')^2 = 6 \times \left(\frac{1}{5}\right)^2 \times \left(\frac{4}{5}\right)^2 = 6 \times \frac{1}{25} \times \frac{16}{25} = \frac{96}{625}\).
- For \(X = 3\) (three defective bulbs): Three bulbs are defective, and one is non-defective. There are \(\binom{4}{3}\) ways this can happen.
\(P(X = 3) = \binom{4}{3} P(D)^3 P(D')^1 = 4 \times \left(\frac{1}{5}\right)^3 \times \frac{4}{5} = 4 \times \frac{1}{125} \times \frac{4}{5} = \frac{16}{625}\).
- For \(X = 4\) (four defective bulbs): All 4 bulbs are defective.
\(P(X = 4) = (P(D))^4 = \left(\frac{1}{5}\right)^4 = \frac{1}{625}\).
The probability distribution for X is:

X	0	1	2	3	4
P(X)	\(\frac{256}{625}\)	\(\frac{256}{625}\)	\(\frac{96}{625}\)	\(\frac{16}{625}\)	\(\frac{1}{625}\)

In simple words: We have 30 bulbs, with 6 being faulty. We pick 4 bulbs, putting each back after checking. X is the count of faulty bulbs we get. We calculate the chance of getting 0, 1, 2, 3, or 4 faulty bulbs and put these chances in a table.

🎯 Exam Tip: For problems involving 'with replacement' selections, use the binomial probability formula \(\binom{n}{k} p^k (1-p)^{n-k}\), where n is the number of trials (sample size), k is the number of successes (defective bulbs), and p is the probability of success in one trial. Ensure you calculate each probability accurately.

 

Question 7. એક સિક્કો અસમતોલ છે. તેને ઉછાળતાં છાપ મળવાની સંભાવના તે કાંટો મળે તેની સંભાવના કરતાં ત્રણ ગણી છે. જો સિક્કાને બે વાર ઉછાળવામાં આવે, તો કાંટાની સંખ્યા માટેનું સંભાવના વિતરણ શોધો.
Answer: A coin is biased.
Let the probability of getting a tail be \(q\).
The probability of getting a head is three times the probability of getting a tail, so \(P(Head) = 3q\).
Since the sum of probabilities for all outcomes must be 1:
\(P(Head) + P(Tail) = 1\)
\(3q + q = 1\)
\(4q = 1\)
\(q = \frac{1}{4}\)
So, the probability of getting a tail is \(P(Tail) = \frac{1}{4}\).
And the probability of getting a head is \(P(Head) = 3 \times \frac{1}{4} = \frac{3}{4}\).
The coin is tossed two times. Let X be the number of tails. The possible values for X are 0, 1, and 2.
- For \(X = 0\) (no tails): This means two heads (HH).
\(P(X = 0) = P(HH) = P(Head) \times P(Head) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\).
- For \(X = 1\) (one tail): This means one tail and one head (TH or HT).
\(P(X = 1) = P(TH) + P(HT) = \left(\frac{1}{4} \times \frac{3}{4}\right) + \left(\frac{3}{4} \times \frac{1}{4}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16}\).
- For \(X = 2\) (two tails): This means two tails (TT).
\(P(X = 2) = P(TT) = P(Tail) \times P(Tail) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\).
The probability distribution of the random variable X is:

X	0	1	2
P(X)	\(\frac{9}{16}\)	\(\frac{6}{16}\)	\(\frac{1}{16}\)

In simple words: We have a special coin where heads are three times more likely than tails. We figure out the exact chance of getting a head (3/4) or a tail (1/4). Then, we flip the coin twice and count how many tails we get (0, 1, or 2). We then write down the probability for each count of tails.

🎯 Exam Tip: For biased coin problems, first determine the individual probabilities of heads and tails. Then, treat each toss as independent for calculating probabilities of combinations like HH, HT, TH, TT. Ensure the sum of P(Head) and P(Tail) equals 1.

 

Question 8. એક યાદચ્છિક ચલ X નું સંભાવના વિતરણ નીચે પ્રમાણે છે :

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k²	2k²	7k² + k

મૂલ્ય નક્કી કરી :
(i) k
(ii) P(X < 3)
(iii) P(X > 6)
(iv) P(0 < x < 3)
Answer:
(i) For any probability distribution, the sum of all probabilities must be 1.
\(\sum_{i=0}^{7} P(X=x_i) = 1\)
\(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 1\)
\(0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1\)
\(10k^2 + 9k - 1 = 0\)
We can factor this quadratic equation:
\(10k^2 + 10k - k - 1 = 0\)
\(10k(k + 1) - 1(k + 1) = 0\)
\((k + 1)(10k - 1) = 0\)
This gives two possible values for k: \(k = -1\) or \(k = \frac{1}{10}\).
Since probabilities cannot be negative, and if \(k = -1\), \(P(X=1) = -1\), which is not possible.
Therefore, \(k = \frac{1}{10}\).
(ii) We need to find \(P(X < 3)\).
\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)
\(P(X < 3) = 0 + k + 2k = 3k\)
Since \(k = \frac{1}{10}\),
\(P(X < 3) = 3 \times \frac{1}{10} = \frac{3}{10}\).
(iii) We need to find \(P(X > 6)\).
\(P(X > 6) = P(X = 7)\)
\(P(X > 6) = 7k^2 + k\)
Substitute \(k = \frac{1}{10}\):
\(P(X > 6) = 7\left(\frac{1}{10}\right)^2 + \frac{1}{10} = 7 \times \frac{1}{100} + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100}\).
(iv) We need to find \(P(0 < X < 3)\).
\(P(0 < X < 3) = P(X = 1) + P(X = 2)\)
\(P(0 < X < 3) = k + 2k = 3k\)
Substitute \(k = \frac{1}{10}\):
\(P(0 < X < 3) = 3 \times \frac{1}{10} = \frac{3}{10}\).In simple words: First, we use the rule that all probabilities must add up to 1 to find the value of 'k'. We solve the equation for 'k' and pick the positive value. Then, we use this 'k' to find the probabilities for different conditions like X being less than 3, greater than 6, or between 0 and 3.

🎯 Exam Tip: Always start by ensuring the sum of all probabilities is equal to 1 to find any unknown constants like 'k'. Remember that individual probabilities must be non-negative. For range-based probabilities like P(X < 3), sum the probabilities of all discrete values within that range.

 

Question 9. યાદૈચ્છિક ચલ Xનું સંભાવના વિતરણ P(X) નીચે આપેલ સ્વરૂપનું છે. k કોઈક સંખ્યા છે : \[P(X) = \begin{cases} k, & x=0 \\ 2k, & x=1 \\ 3k, & x=2 \\ 0, & \text{otherwise} \end{cases}\]
(a) k નું મૂલ્ય શોધો.
(b) P(X < 2), P(X ≤ 2), P(X ≥ 2) શોધો.
Answer: The probability distribution for the random variable X is given as:

X	0	1	2
P(X)	k	2k	3k

(a) To find the value of k, we use the property that the sum of all probabilities must be 1.
\(P(X = 0) + P(X = 1) + P(X = 2) = 1\)
\(k + 2k + 3k = 1\)
\(6k = 1\)
\(k = \frac{1}{6}\).
(b) We need to find \(P(X < 2)\), \(P(X \leq 2)\), and \(P(X \geq 2)\).
- For \(P(X < 2)\): This means \(P(X = 0) + P(X = 1)\).
\(P(X < 2) = k + 2k = 3k\)
Substitute \(k = \frac{1}{6}\):
\(P(X < 2) = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\).
- For \(P(X \leq 2)\): This means \(P(X = 0) + P(X = 1) + P(X = 2)\).
\(P(X \leq 2) = k + 2k + 3k = 6k\)
Substitute \(k = \frac{1}{6}\):
\(P(X \leq 2) = 6 \times \frac{1}{6} = 1\).
- For \(P(X \geq 2)\): This means \(P(X = 2)\).
\(P(X \geq 2) = 3k\)
Substitute \(k = \frac{1}{6}\):
\(P(X \geq 2) = 3 \times \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\).In simple words: First, we find 'k' by adding all the probabilities and setting them equal to 1. Once 'k' is known, we can calculate probabilities for X being less than 2, less than or equal to 2, or greater than or equal to 2 by adding the probabilities of the specific X values.

🎯 Exam Tip: Always ensure the sum of all probabilities in a distribution is 1 to solve for unknown constants. For inequalities, carefully determine which discrete values of X are included in the specified range (e.g., \(X < 2\) includes X=0 and X=1, but not X=2).

 

Question 10. એક સમતોલ સિક્કાને ત્રણ વખત ઉછાળતાં મળતી છાપની સંખ્યાનો મધ્યક શોધો.
Answer: A fair coin is tossed three times.
The total number of possible outcomes is \(n(S) = 2^3 = 8\).
Let X be the number of heads. The possible values for X are 0, 1, 2, and 3.
- For \(X = 0\) (no heads): This occurs with outcome TTT. So, \(P(X = 0) = P(TTT) = \frac{1}{8}\).
- For \(X = 1\) (one head): This occurs with outcomes HTT, THT, TTH. So, \(P(X = 1) = \frac{3}{8}\).
- For \(X = 2\) (two heads): This occurs with outcomes HHT, HTH, THH. So, \(P(X = 2) = \frac{3}{8}\).
- For \(X = 3\) (three heads): This occurs with outcome HHH. So, \(P(X = 3) = \frac{1}{8}\).
The probability distribution of X is:

X	0	1	2	3
P(X)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)

The mean (mathematical expectation) \(E(X)\) is calculated as \(\sum x_i p(x_i)\).
\(E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))\)
\(E(X) = \left(0 \times \frac{1}{8}\right) + \left(1 \times \frac{3}{8}\right) + \left(2 \times \frac{3}{8}\right) + \left(3 \times \frac{1}{8}\right)\)
\(E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8}\)
\(E(X) = \frac{3 + 6 + 3}{8} = \frac{12}{8} = \frac{3}{2}\).In simple words: We toss a fair coin three times and count how many heads appear. We find the probability for each possible number of heads (0, 1, 2, or 3). To find the mean, we multiply each number of heads by its probability and add all these results together.

🎯 Exam Tip: For calculating the mean of a discrete probability distribution, ensure all probabilities sum to 1. The formula \(E(X) = \sum x_i P(X=x_i)\) is fundamental; make sure to multiply each value of the random variable by its corresponding probability before summing.

 

Question 11. બે પાસાને એક સાથે ફેંકવામાં આવે છે. જો X 6 મળવાની કુલ સંખ્યા દર્શાવે નો Xની ગાણિતિક અપેક્ષા શોધી
Answer: A pair of dice is thrown simultaneously.
Let X represent the total number of 6s obtained.
When a single die is thrown, the probability of getting a 6 is \(P(6) = \frac{1}{6}\).
The probability of not getting a 6 is \(P(\text{not } 6) = 1 - \frac{1}{6} = \frac{5}{6}\).
The possible values for X (number of 6s on two dice) are 0, 1, and 2.
- For \(X = 0\): Neither die shows a 6.
\(P(X = 0) = P(\text{not } 6) \times P(\text{not } 6) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\).
- For \(X = 1\): Exactly one die shows a 6.
\(P(X = 1) = (P(6) \times P(\text{not } 6)) + (P(\text{not } 6) \times P(6)) = \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36}\).
- For \(X = 2\): Both dice show a 6.
\(P(X = 2) = P(6) \times P(6) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
The probability distribution of X is:

X	0	1	2
P(X)	\(\frac{25}{36}\)	\(\frac{10}{36}\)	\(\frac{1}{36}\)

The mathematical expectation (mean) \(E(X)\) is calculated as \(\sum x_i p(x_i)\).
\(E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))\)
\(E(X) = \left(0 \times \frac{25}{36}\right) + \left(1 \times \frac{10}{36}\right) + \left(2 \times \frac{1}{36}\right)\)
\(E(X) = 0 + \frac{10}{36} + \frac{2}{36}\)
\(E(X) = \frac{10 + 2}{36} = \frac{12}{36} = \frac{1}{3}\).In simple words: When two dice are thrown, X is the number of times we get a '6'. We find the probability of getting zero, one, or two '6's. Then, to find the average number of '6's (expectation), we multiply each number of '6's by its probability and sum them up.

🎯 Exam Tip: Clearly define the random variable X and its possible values. For dice problems, listing individual probabilities (like P(6) and P(not 6)) first simplifies calculating probabilities for combinations. The mean \(E(X)\) represents the average outcome over many trials.

 

Question 12. પ્રથમ છ ધન પૂર્ણાંકોમાંથી યાદચ્છિક રીતે બે સંખ્યાઓ પસંદ (પુરવણીરહિત) કરી છે. ધારો કે જો X હોળવેલી સંખ્યાઓ પૈકી મોટી સંખ્યા દર્શાવે છે. E(X) શોધો.
Answer: Two numbers are randomly selected without replacement from the first six positive integers \(\{1, 2, 3, 4, 5, 6\}\).
The total number of ordered pairs (without replacement) that can be selected is \(P(6,2) = 6 \times 5 = 30\).
Let X denote the larger of the two selected numbers.
The possible values for X are 2, 3, 4, 5, 6.
- For \(X = 2\): The pairs where 2 is the larger number are \((1, 2), (2, 1)\).
\(P(X = 2) = \frac{2}{30}\).
- For \(X = 3\): The pairs where 3 is the larger number are \((1, 3), (3, 1), (2, 3), (3, 2)\).
\(P(X = 3) = \frac{4}{30}\).
- For \(X = 4\): The pairs where 4 is the larger number are \((1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3)\).
\(P(X = 4) = \frac{6}{30}\).
- For \(X = 5\): The pairs where 5 is the larger number are \((1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4)\).
\(P(X = 5) = \frac{8}{30}\).
- For \(X = 6\): The pairs where 6 is the larger number are \((1, 6), (6, 1), (2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (5, 6), (6, 5)\).
\(P(X = 6) = \frac{10}{30}\).
The probability distribution of X is:

X	2	3	4	5	6
P(X)	\(\frac{2}{30}\)	\(\frac{4}{30}\)	\(\frac{6}{30}\)	\(\frac{8}{30}\)	\(\frac{10}{30}\)

The mathematical expectation (mean) \(E(X)\) is calculated as \(\sum x_i p(x_i)\).
\(E(X) = \left(2 \times \frac{2}{30}\right) + \left(3 \times \frac{4}{30}\right) + \left(4 \times \frac{6}{30}\right) + \left(5 \times \frac{8}{30}\right) + \left(6 \times \frac{10}{30}\right)\)
\(E(X) = \frac{4}{30} + \frac{12}{30} + \frac{24}{30} + \frac{40}{30} + \frac{60}{30}\)
\(E(X) = \frac{4 + 12 + 24 + 40 + 60}{30} = \frac{140}{30} = \frac{14}{3}\).In simple words: We pick two different numbers from 1 to 6. X is the bigger number we picked. We find how many pairs result in X being 2, 3, 4, 5, or 6. We then calculate the chance for each X value. Finally, to find the average value of X, we multiply each X by its probability and add them up.

🎯 Exam Tip: When selecting items without replacement, consider ordered pairs if the problem implies distinct sequences or if you are counting specific events (like which number is larger). Carefully list all outcomes for each value of the random variable X to correctly determine its probabilities. Ensure the sum of probabilities equals 1.

 

Question 13. ધારો કે X એ બે સમતોલ પાસાઓને ઉછાળતાં મળતી સંખ્યાઓનો સરવાળો દર્શાવે છે. તો Xનું વિચરણ અને પ્રમાણિત વિચલન શોધો.
Answer: Let X denote the sum of the numbers obtained when two fair dice are thrown.
When two fair dice are thrown, the total number of possible outcomes is \(n(S) = 6 \times 6 = 36\).
The possible values for X (the sum) range from 2 to 12.
The probability distribution of X is:
\(P(X=2) = P(\{1,1\}) = \frac{1}{36}\)
\(P(X=3) = P(\{1,2\}, \{2,1\}) = \frac{2}{36}\)
\(P(X=4) = P(\{1,3\}, \{2,2\}, \{3,1\}) = \frac{3}{36}\)
\(P(X=5) = P(\{1,4\}, \{2,3\}, \{3,2\}, \{4,1\}) = \frac{4}{36}\)
\(P(X=6) = P(\{1,5\}, \{2,4\}, \{3,3\}, \{4,2\}, \{5,1\}) = \frac{5}{36}\)
\(P(X=7) = P(\{1,6\}, \{2,5\}, \{3,4\}, \{4,3\}, \{5,2\}, \{6,1\}) = \frac{6}{36}\)
\(P(X=8) = P(\{2,6\}, \{3,5\}, \{4,4\}, \{5,3\}, \{6,2\}) = \frac{5}{36}\)
\(P(X=9) = P(\{3,6\}, \{4,5\}, \{5,4\}, \{6,3\}) = \frac{4}{36}\)
\(P(X=10) = P(\{4,6\}, \{5,5\}, \{6,4\}) = \frac{3}{36}\)
\(P(X=11) = P(\{5,6\}, \{6,5\}) = \frac{2}{36}\)
\(P(X=12) = P(\{6,6\}) = \frac{1}{36}\)
The probability distribution table for X is:

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{4}{36}\)	\(\frac{5}{36}\)	\(\frac{6}{36}\)	\(\frac{5}{36}\)	\(\frac{4}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{1}{36}\)

First, calculate the Mean \(E(X) = \sum x_i p(x_i)\).
\(E(X) = \frac{1}{36} [ (2 \times 1) + (3 \times 2) + (4 \times 3) + (5 \times 4) + (6 \times 5) + (7 \times 6) + (8 \times 5) + (9 \times 4) + (10 \times 3) + (11 \times 2) + (12 \times 1) ]\)
\(E(X) = \frac{1}{36} [ 2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 ]\)
\(E(X) = \frac{252}{36} = 7\).
Next, calculate \(E(X^2) = \sum x_i^2 p(x_i)\).
\(E(X^2) = \frac{1}{36} [ (2^2 \times 1) + (3^2 \times 2) + (4^2 \times 3) + (5^2 \times 4) + (6^2 \times 5) + (7^2 \times 6) + (8^2 \times 5) + (9^2 \times 4) + (10^2 \times 3) + (11^2 \times 2) + (12^2 \times 1) ]\)
\(E(X^2) = \frac{1}{36} [ (4 \times 1) + (9 \times 2) + (16 \times 3) + (25 \times 4) + (36 \times 5) + (49 \times 6) + (64 \times 5) + (81 \times 4) + (100 \times 3) + (121 \times 2) + (144 \times 1) ]\)
\(E(X^2) = \frac{1}{36} [ 4 + 18 + 48 + 100 + 180 + 294 + 320 + 324 + 300 + 242 + 144 ]\)
\(E(X^2) = \frac{1974}{36} = \frac{329}{6}\).
Now, calculate the Variance \(Var(X) = E(X^2) - [E(X)]^2\).
\(Var(X) = \frac{329}{6} - (7)^2\)
\(Var(X) = 54.83 - 49 = 5.83\).
Finally, calculate the Standard Deviation \(\sigma_X = \sqrt{Var(X)}\).
\(\sigma_X = \sqrt{5.83} \approx 2.4\).In simple words: When two dice are thrown, X is the sum of the numbers. We list all possible sums (2 to 12) and their chances. Then, we find the average sum (mean). Next, we calculate the average of the squared sums. Using these two averages, we find how spread out the sums are (variance) and then its square root (standard deviation).

🎯 Exam Tip: For problems requiring variance and standard deviation, accurately determine the probability distribution first. Remember the formulas: \(E(X) = \sum x_i P(X=x_i)\) and \(Var(X) = E(X^2) - [E(X)]^2\). Listing all outcomes for dice sums can prevent errors in probability calculation. \(E(X^2)\) requires summing \(x_i^2 P(X=x_i)\).

 

Question 14. એક વર્ગમાં 15 વિદ્યાર્થીઓ છે. તેમની વય 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 અને 20 વર્ષ છે. એક વિધાર્થી પસંદ કરવામાં આવ્યો છે. પ્રત્યેક વિધાર્થી પસંદ થવાની સમાન સંભાવના હતી અને પસંદ થયેલા વિદ્યાર્થીની વય X નોંધી છે. યાદૈચ્છિક ચલ Xનું સંભાવના વિતરણ શું છે ? Xનો મધ્યક, વિચરણ અને પ્રમાણિત વિચલન શોધો.
Answer: There are 15 students in a class. Their ages are: 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, 20.
A student is selected randomly, and X is the age of the selected student. Each student has an equal probability of being selected, which is \(\frac{1}{15}\).
First, let's count the frequency of each age to find the probability distribution:
- Age 14: Appears 2 times. \(P(X=14) = \frac{2}{15}\)
- Age 15: Appears 1 time. \(P(X=15) = \frac{1}{15}\)
- Age 16: Appears 2 times. \(P(X=16) = \frac{2}{15}\)
- Age 17: Appears 3 times. \(P(X=17) = \frac{3}{15}\)
- Age 18: Appears 1 time. \(P(X=18) = \frac{1}{15}\)
- Age 19: Appears 2 times. \(P(X=19) = \frac{2}{15}\)
- Age 20: Appears 3 times. \(P(X=20) = \frac{3}{15}\)
- Age 21: Appears 1 time. \(P(X=21) = \frac{1}{15}\)
The probability distribution of the random variable X is:

X (Age)	Frequency	P(X)	\(x_i P(x_i)\)	\(x_i^2 P(x_i)\)
14	2	\(\frac{2}{15}\)	\(\frac{28}{15}\)	\(\frac{392}{15}\)
15	1	\(\frac{1}{15}\)	\(\frac{15}{15}\)	\(\frac{225}{15}\)
16	2	\(\frac{2}{15}\)	\(\frac{32}{15}\)	\(\frac{512}{15}\)
17	3	\(\frac{3}{15}\)	\(\frac{51}{15}\)	\(\frac{867}{15}\)
18	1	\(\frac{1}{15}\)	\(\frac{18}{15}\)	\(\frac{324}{15}\)
19	2	\(\frac{2}{15}\)	\(\frac{38}{15}\)	\(\frac{722}{15}\)
20	3	\(\frac{3}{15}\)	\(\frac{60}{15}\)	\(\frac{1200}{15}\)
21	1	\(\frac{1}{15}\)	\(\frac{21}{15}\)	\(\frac{441}{15}\)
		Sum = 15	\(\sum x_i P(x_i) = \frac{263}{15}\)	\(\sum x_i^2 P(x_i) = \frac{4683}{15}\)

Mean (\(\mu\)) \(E(X) = \sum x_i P(x_i) = \frac{263}{15} \approx 17.53\).
Variance \(Var(X) = \sum x_i^2 P(x_i) - [E(X)]^2\)
\(Var(X) = \frac{4683}{15} - \left(\frac{263}{15}\right)^2\)
\(Var(X) = 312.20 - (17.53)^2\)
\(Var(X) = 312.20 - 307.30 = 4.9\).
Standard Deviation (\(\sigma_X\)) \( = \sqrt{Var(X)}\)
\(\sigma_X = \sqrt{4.9} \approx 2.2\).In simple words: We have a class of 15 students with different ages. We pick one student randomly. X is the age of that student. We count how many students are each age to find the chance of picking each age. Then, we calculate the average age (mean), how much the ages vary (variance), and the typical spread of ages (standard deviation).

🎯 Exam Tip: For distribution from raw data, first organize the data into a frequency distribution to derive the probabilities \(P(X=x_i)\). Carefully calculate \(\sum x_i P(x_i)\) for the mean and \(\sum x_i^2 P(x_i)\) for \(E(X^2)\) to avoid errors in variance and standard deviation calculations. Accuracy in arithmetic is crucial.

 

Question 14. એક વર્ગમાં 15 વિદ્યાર્થીઓ છે. તેમની વય 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 અને 20 વર્ષ છે. એક વિધાર્થી પસંદ કરવામાં આવ્યો છે. પ્રત્યેક વિધાર્થી પસંદ થવાની સમાન સંભાવના હતી અને પસંદ થયેલા વિદ્યાર્થીની વય X નોંધી છે. યાદૈચ્છિક ચલ Xનું સંભાવના વિતરણ શું છે ? Xનો મધ્યક, વિચરણ અને પ્રમાણિત વિચલન શોધો.
Answer:In a class, there are 15 students. Their ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, and 20 years. One student is chosen randomly. Each student has an equal chance of being selected. The age of the chosen student is noted as X. The total number of students in the class is 15. So, \(n(S) = 15\). The random variable X represents the age of the chosen student. The probability distribution for X is as follows:

X (વય)	X ધરાવતાં વિદ્યાર્થીની સંખ્યા	P(X)	\(x_i p(x_i)\)	\(x_i^2 p(x_i)\)
14	2	\(\frac{2}{15}\)	\(\frac{28}{15}\)	\(\frac{392}{15}\)
15	1	\(\frac{1}{15}\)	\(\frac{15}{15}\)	\(\frac{225}{15}\)
16	2	\(\frac{2}{15}\)	\(\frac{32}{15}\)	\(\frac{512}{15}\)
17	3	\(\frac{3}{15}\)	\(\frac{51}{15}\)	\(\frac{867}{15}\)
18	1	\(\frac{1}{15}\)	\(\frac{18}{15}\)	\(\frac{324}{15}\)
19	2	\(\frac{2}{15}\)	\(\frac{38}{15}\)	\(\frac{722}{15}\)
20	3	\(\frac{3}{15}\)	\(\frac{60}{15}\)	\(\frac{1200}{15}\)
21	1	\(\frac{1}{15}\)	\(\frac{21}{15}\)	\(\frac{441}{15}\)
\(\sum x_i p(x_i)\)			\(\frac{263}{15}\)	\(\frac{4683}{15}\)

We can find the mean, variance, and standard deviation. The mean \( \mu \) is given by \( \sum x_i p(x_i) \). \[ \mu = E(X) = \sum x_i p(x_i) \] \[ = \frac{263}{15} \] \[ = 17.53 \] The variance \( Var(X) \) is given by \( \sum x_i^2 p(x_i) - [E(X)]^2 \). \[ Var(X) = \sum x_i^2 p(x_i) - [E(X)]^2 \] \[ = \frac{4683}{15} - \left(\frac{263}{15}\right)^2 \] \[ = 312.20 - (17.53)^2 \] \[ = 312.20 - 307.30 \] \[ = 4.9 \] The standard deviation \( \sigma_X \) is the square root of the variance. \[ \sigma_X = \sqrt{Var(X)} \] \[ = \sqrt{4.9} \] \[ = 2.2 \]In simple words: We find the probability for each age, then calculate the average age (mean), how much the ages spread out (variance), and the typical deviation from the average (standard deviation).

🎯 Exam Tip: Remember to list all possible ages (X values) and their corresponding probabilities P(X) to form the probability distribution. Mean and variance calculations require accurate summation of \(x_i p(x_i)\) and \(x_i^2 p(x_i)\).

 

Question 15. એક બેઠકમાં, એક નિશ્ચિત દરખાસ્તની તરફેણમાં 70 % સભ્યો અને તેની વિરોધમાં 30 % સભ્યો છે. એક સભ્ય યાદચ્છિક રીતે પસંદ કર્યો અને જો તે વિરોધ કરે, તો આપણે X = 0 અને જો તે તરફેણમાં હોય તો X = 1 લઈએ. E(X) અને Var(X) શોધો.
Answer:In a meeting, 70% of members support a proposal, and 30% oppose it. A member is chosen randomly. If the member opposes the proposal, X = 0. If the member supports it, X = 1. We need to find E(X) and Var(X). The probability that a member opposes the proposal (X = 0) is 30%. \[ P(X = 0) = P(\text{विरोध કરે}) = \frac{30}{100} = \frac{3}{10} \] The probability that a member supports the proposal (X = 1) is 70%. \[ P(X = 1) = P(\text{તરફેણ કરે}) = \frac{70}{100} = \frac{7}{10} \] The probability distribution is:

X	P(X)	\(x_i p(x_i)\)	\(x_i^2 p(x_i)\)
0	\(\frac{3}{10}\)	0	0
1	\(\frac{7}{10}\)	\(\frac{7}{10}\)	\(\frac{7}{10}\)

We calculate the expected value E(X). \[ E(X) = \sum x_i p(x_i) \] \[ = 0 + \frac{7}{10} \] \[ = 0.7 \] Next, we calculate the variance Var(X). \[ Var(X) = \sum x_i^2 p(x_i) - [E(X)]^2 \] \[ = \left(0 + \frac{7}{10}\right) - (0.7)^2 \] \[ = 0.7 - 0.49 \] \[ = 0.21 \]In simple words: We calculate the average outcome (expected value) by multiplying each possible outcome (X) by its chance (P(X)) and adding them up. Then we find the variance, which shows how much the actual outcomes differ from this average.

🎯 Exam Tip: For binary random variables like this (0 or 1), remember that \(E(X) = P(X=1)\) and \(Var(X) = P(X=1) \cdot P(X=0)\) can simplify calculations.

 

Question 16. એક પાસાના ત્રણ પૃષ્ઠ પર 1, બે પૃષ્ઠ પર 2 અને એક પર 5 અંકિત હોય, તો તેને ઉછાળતાં મળતી સંખ્યાઓનો મધ્યક ............... છે.
(A) 1
(B) 2
(C) 5
(D) \(\frac{8}{3}\)
Answer: (B) 2When a die has three faces marked 1, two faces marked 2, and one face marked 5, we want to find the mean of the numbers obtained when it is rolled. The possible outcomes are X: 1, 2, 5. The probabilities are: For X = 1: \( P(X=1) = \frac{3}{6} \) (since 3 faces have 1) For X = 2: \( P(X=2) = \frac{2}{6} \) (since 2 faces have 2) For X = 5: \( P(X=5) = \frac{1}{6} \) (since 1 face has 5) The probability distribution is:

X	P(X)	\(x_i p(x_i)\)
1	\(\frac{3}{6}\)	\(\frac{3}{6}\)
2	\(\frac{2}{6}\)	\(\frac{4}{6}\)
5	\(\frac{1}{6}\)	\(\frac{5}{6}\)
\(\sum x_i p(x_i)\)		\(\frac{12}{6}\)

We calculate the mean E(X). \[ E(X) = \sum x_i p(x_i) \] \[ = \frac{12}{6} \] \[ = 2 \] Therefore, option (B) is the correct answer.In simple words: To find the average number rolled on this special die, we multiply each number by its chance of appearing and add these results.

🎯 Exam Tip: For such problems, first determine the probability of each outcome, then use the formula for expected value \(E(X) = \sum x_i P(X=x_i)\).

 

Question 17. ધારો કે પત્તાંની થોકડીમાંથી બે પત્તાં યાદચ્છિક રીતે પસંદ કરવામાં આવે છે. ધારો કે મેં એ મળેલ એક્કાઓની સંખ્યા દર્શાવે છે, તો E(X) નું મૂલ્ય ............... . . .
(A) \(\frac{37}{221}\)
(B) \(\frac{5}{13}\)
(C) \(\frac{1}{13}\)
(D) \(\frac{2}{13}\)
Answer: (D) \(\frac{2}{13}\)Suppose two cards are drawn randomly from a deck of cards. Let X represent the number of aces obtained. We need to find the value of E(X). A standard deck has 52 cards, including 4 aces. When two cards are drawn, the total number of ways to draw 2 cards from 52 is \( {52 \choose 2} \). \[ {52 \choose 2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \] The random variable X can take values 0, 1, or 2, representing the number of aces drawn. If X = 0 (no aces are drawn): This means 2 cards are drawn from the 48 non-ace cards. \[ P(X = 0) = \frac{{48 \choose 2}}{{52 \choose 2}} = \frac{\frac{48 \times 47}{2 \times 1}}{\frac{52 \times 51}{2 \times 1}} = \frac{48 \times 47}{52 \times 51} = \frac{2256}{2652} = \frac{188}{221} \] If X = 1 (one ace is drawn): This means 1 ace is drawn from 4 aces and 1 non-ace is drawn from 48 non-ace cards. \[ P(X = 1) = \frac{{4 \choose 1} \times {48 \choose 1}}{{52 \choose 2}} = \frac{4 \times 48}{1326} = \frac{192}{1326} = \frac{32}{221} \] If X = 2 (two aces are drawn): This means 2 aces are drawn from 4 aces. \[ P(X = 2) = \frac{{4 \choose 2}}{{52 \choose 2}} = \frac{\frac{4 \times 3}{2 \times 1}}{1326} = \frac{6}{1326} = \frac{1}{221} \] The probability distribution for X is:

X	P(X)
0	\(\frac{188}{221}\)
1	\(\frac{32}{221}\)
2	\(\frac{1}{221}\)

Now, we calculate the expected value E(X). \[ E(X) = \sum x_i p(x_i) \] \[ = \left(0 \times \frac{188}{221}\right) + \left(1 \times \frac{32}{221}\right) + \left(2 \times \frac{1}{221}\right) \] \[ = 0 + \frac{32}{221} + \frac{2}{221} \] \[ = \frac{34}{221} \] \[ = \frac{2}{13} \] Therefore, option (D) is the correct answer.In simple words: We figure out the chance of getting zero, one, or two aces when picking two cards. Then, we multiply each number of aces by its chance and add them up to find the average number of aces we expect to get.

🎯 Exam Tip: When drawing cards without replacement, use combinations \({n \choose k}\) to calculate probabilities. Ensure the sum of probabilities for all possible outcomes equals 1.