JEE Mathematics Probability MCQs Set 28

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Multiple Choice Questions

Question 1: A natural number \(x\) is chosen at random from the first 100 natural numbers. The probability that \(x+\dfrac{100}x>50\) is
(a) \(\dfrac{1}{20}\)
(b) \(\dfrac{1}{3}\)
(c) \(\dfrac{11}{20}\)
(d) \(\dfrac{3}{20}\)
Show Answer & Explanation

Answer: (c) \(\dfrac{11}{20}\)

Explanation:
1. \(x^2-50x+100>0\) has roots \(25\pm\sqrt{525}\approx2.08,\ 47.9\).
2. So \(x=1,2\) or \(x=48,\dots,100\): \(2+53=55\) values, probability \(\dfrac{55}{100}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 2: A person throws two dice, one a regular cube and the other a regular tetrahedron with numbers 1, 2, 3, 4 on its faces, the number on the lowest face being taken for the tetrahedron. The chance that the sum of numbers thrown is not less than 5 is
(a) \(\dfrac{1}{4}\)
(b) \(\dfrac{4}{5}\)
(c) \(\dfrac{5}{6}\)
(d) \(\dfrac{3}{4}\)
Show Answer & Explanation

Answer: (d) \(\dfrac{3}{4}\)

Explanation:
1. Total outcomes \(6\times4=24\).
2. For tetrahedron number \(t=1,2,3,4\) the cube needs \(\ge4,3,2,1\): \(3+4+5+6=18\) outcomes; \(\dfrac{18}{24}=\dfrac34\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 3: \(A\) is a set containing \(n\) elements. A subset \(P\) of \(A\) is chosen at random, and the elements of \(P\) are replaced in \(A\). Another subset \(Q\) of \(A\) is then chosen at random. The probability that \(P\cup Q\) contains exactly \(r\) elements, \(1\le r\le n\), is
(a) \(\dfrac{{}^nC_r\,3^r}{4^n}\)
(b) \(\dfrac{3^n}{4^n}\)
(c) \(\dfrac{{}^nC_r\,3^{n-r}}{4^n}\)
(d) \(\dfrac{{}^nC_r\,3^n}{4^n}\)
Show Answer & Explanation

Answer: (a) \(\dfrac{{}^nC_r\,3^r}{4^n}\)

Explanation:
1. Each element lies in \(P\) or \(Q\) (or both) with probability \(1-\dfrac14=\dfrac34\), independently.
2. Exactly \(r\) elements: \({}^nC_r\left(\dfrac34\right)^r\left(\dfrac14\right)^{n-r}=\dfrac{{}^nC_r\,3^r}{4^n}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 4: If the letters of the word MATHEMATICS are arranged arbitrarily, the probability that C comes before E, E before H, H before I and I before S is
(a) \(\dfrac{1}{75}\)
(b) \(\dfrac{1}{24}\)
(c) \(\dfrac{1}{120}\)
(d) \(\dfrac{1}{720}\)
Show Answer & Explanation

Answer: (c) \(\dfrac{1}{120}\)

Explanation:
1. The five letters C, E, H, I, S each occur once; among their \(5!=120\) relative orders exactly one is the required order.
2. Probability \(=\dfrac1{120}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 5: In a game of chance a player throws a pair of dice and scores points equal to the difference between the numbers on the two dice. The winner is the person who scores exactly 5 points more than his opponent. If two players play this game only once, the probability that neither of them wins is
(a) \(\dfrac{1}{54}\)
(b) \(\dfrac{53}{54}\)
(c) \(\dfrac{1}{108}\)
(d) \(\dfrac{107}{108}\)
Show Answer & Explanation

Answer: (b) \(\dfrac{53}{54}\)

Explanation:
1. Score 0 has probability \(\dfrac6{36}=\dfrac16\); score 5 has probability \(\dfrac2{36}=\dfrac1{18}\).
2. A win needs scores \((5,0)\) or \((0,5)\): \(2\cdot\dfrac16\cdot\dfrac1{18}=\dfrac1{54}\).
3. Neither wins: \(1-\dfrac1{54}=\dfrac{53}{54}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 6: A box contains 2 fifty-paise coins, 5 twenty-five-paise coins and 15 ten-paise coins. Five coins are taken out of the box at random. The probability that the value of these five coins is less than one rupee and fifty paise is
(a) \(1-\dfrac{150}{{}^{22}C_5}\)
(b) \(\dfrac{170}{{}^{22}C_5}\)
(c) \(1-\dfrac{170}{{}^{22}C_5}\)
(d) none of these
Show Answer & Explanation

Answer: (c) \(1-\dfrac{170}{{}^{22}C_5}\)

Explanation:
1. Total \({}^{22}C_5=26334\). A value of 150 paise or more arises only from: two 50p and three 25p (175 paise), two 50p, two 25p and one 10p (160 paise), or one 50p and four 25p (150 paise).
2. Counting: \({}^5C_3=10\), \({}^5C_2\cdot15=150\), \(2\cdot{}^5C_4=10\): total \(170\).
3. Probability of less than Rs. 1.50 is \(1-\dfrac{170}{{}^{22}C_5}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 7: There are two vans, each having numbered seats, 3 in the front and 4 at the back. There are 3 girls and 9 boys to be seated in the vans. The probability that the 3 girls sit together in a back row on adjacent seats is
(a) \(\dfrac{1}{13}\)
(b) \(\dfrac{1}{91}\)
(c) \(\dfrac{1}{39}\)
(d) \(\dfrac{1}{65}\)
Show Answer & Explanation

Answer: (b) \(\dfrac{1}{91}\)

Explanation:
1. The 12 persons occupy 12 of the 14 seats: total arrangements \({}^{14}P_{12}=\dfrac{14!}{2!}\).
2. Adjacent triples in a back row: \(2\times2=4\); girls arranged in \(3!\) ways; boys in the other 11 seats: \({}^{11}P_9=\dfrac{11!}{2!}\).
3. Probability \(=\dfrac{4\cdot6\cdot11!}{14!}=\dfrac{24}{2184}=\dfrac1{91}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 8: For three events \(A,B,C\): \(P(\text{exactly one of }A\text{ or }B)=P(\text{exactly one of }B\text{ or }C)=P(\text{exactly one of }C\text{ or }A)=p\) and \(P(\text{all three occur})=p^2\), where \(0<p<\dfrac12\). The probability of at least one of the three events occurring is
(a) \(\dfrac{p+3p^2}4\)
(b) \(\dfrac{p+3p^2}2\)
(c) \(\dfrac{3p+2p^2}2\)
(d) \(\dfrac{3p+2p^2}4\)
Show Answer & Explanation

Answer: (c) \(\dfrac{3p+2p^2}2\)

Explanation:
1. Adding the three conditions: \(2\sum P(A)-2\sum P(A\cap B)=3p\).
2. \(\sum P(A)-\sum P(A\cap B)=\dfrac{3p}2\).
3. \(P(A\cup B\cup C)=\dfrac{3p}2+p^2=\dfrac{3p+2p^2}2\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 9: There are three events \(E_1,E_2,E_3\), one of which must, and only one can, happen. The odds are 7 to 4 against \(E_1\) and 5 to 3 against \(E_2\). The odds against \(E_3\) are
(a) \(4:11\)
(b) \(3:8\)
(c) \(23:88\)
(d) \(65:23\)
Show Answer & Explanation

Answer: (d) \(65:23\)

Explanation:
1. \(P(E_1)=\dfrac4{11},\ P(E_2)=\dfrac38\).
2. \(P(E_3)=1-\dfrac4{11}-\dfrac38=\dfrac{23}{88}\), so the odds against are \(65:23\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 10: Ashmit and Bishmit appear in an examination. The probability that Ashmit passes is 0.5, the probability that both pass is at least 0.1 and at most 0.3, and the probability that Bishmit passes is \(P(B)\). The range of \(P(B)\) is
(a) \((0.2,\ 0.8)\)
(b) \((0.15,\ 0.25)\)
(c) \([0.3,\ 0.5]\)
(d) \([0.1,\ 0.8]\)
Show Answer & Explanation

Answer: (d) \([0.1,\ 0.8]\)

Explanation:
1. \(P(B)\ge P(A\cap B)\ge0.1\).
2. \(P(A\cup B)\le1\) gives \(P(B)\le1-0.5+P(A\cap B)\le0.8\).
3. So \(0.1\le P(B)\le0.8\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 11: A four-digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is
(a) \(\dfrac13\)
(b) \(\dfrac14\)
(c) \(\dfrac12\)
(d) \(\dfrac34\)
Show Answer & Explanation

Answer: (c) \(\dfrac12\)

Explanation:
1. Total \(4!=24\); the unit digit is 1 or 3 in \(2\cdot3!=12\) of them.
2. Probability \(\dfrac{12}{24}=\dfrac12\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 12: From a well-shuffled pack of 52 cards, four cards are drawn. The probability of drawing exactly 2 spades and exactly 2 aces is
(a) \(\dfrac{1494}{270725}\)
(b) \(\dfrac{1594}{270725}\)
(c) \(\dfrac{1296}{270725}\)
(d) \(\dfrac{1396}{270725}\)
Show Answer & Explanation

Answer: (a) \(\dfrac{1494}{270725}\)

Explanation:
1. Case 1, the ace of spades is drawn (it is both a spade and an ace): one more spade from the 12 non-ace spades, one more ace from the 3 other aces, and one card that is neither: \(12\cdot3\cdot36=1296\).
2. Case 2, the ace of spades is not drawn: two of the 12 non-ace spades and two of the 3 other aces: \({}^{12}C_2\cdot{}^3C_2=66\cdot3=198\).
3. Total \(1296+198=1494\) out of \({}^{52}C_4=270725\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 13: Two numbers \(x\) and \(y\) are chosen at random without replacement from \(1,2,3,\dots,3n\). If the probability that \(x^3+y^3\) is divisible by 3 is \(\dfrac pq\) in lowest terms, then \(p+q\) is
(a) 3
(b) 4
(c) 5
(d) 7
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. \(x^3\equiv x\pmod3\), so we need \(x+y\equiv0\pmod3\): residues \((0,0)\) or \((1,2)\).
2. Favourable \({}^nC_2+n^2=\dfrac{n(3n-1)}2\); total \(\dfrac{3n(3n-1)}2\); the probability is \(\dfrac13\), so \(p+q=4\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 14: 5 girls and 10 boys sit at random in a row having 15 chairs numbered 1 to 15. If the probability that the end seats are occupied by girls and between any two girls an odd number of boys take seats is \(\dfrac{20}n\), then \(\dfrac n{1001}\) is equal to
(a) 2
(b) 4
(c) 3
(d) 6
Show Answer & Explanation

Answer: (c) 3

Explanation:
1. Girls occupy seat positions with both ends included and the 4 gaps each containing an odd number \(2a_i+1\) of boys, summing to 10: \(a_1+a_2+a_3+a_4=3\), giving \({}^6C_3=20\) patterns.
2. Total patterns \({}^{15}C_5=3003\), so \(n=3003\) and \(\dfrac n{1001}=3\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 15: A quadratic equation is chosen from the set of all quadratic equations which are unchanged by squaring their roots. If the chance that the chosen equation has equal roots is \(\dfrac pq\) in lowest terms, then \(p+q\) is
(a) 2
(b) 5
(c) 4
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. Roots \(\alpha,\beta\) with \(\{\alpha^2,\beta^2\}=\{\alpha,\beta\}\): \((0,0),(1,1),(0,1)\), or \((\omega,\omega^2)\) from \(\alpha^2=\beta,\beta^2=\alpha\).
2. Four equations: \(x^2,\ (x-1)^2,\ x(x-1),\ x^2+x+1\); two have equal roots: probability \(\dfrac12\), so \(p+q=3\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 16: A class contains 20 boys and 20 girls of which half the boys and half the girls have cat eyes. If one student is selected from the class, the probability that the student is a boy or has cat eyes is \(\dfrac ab\) in lowest terms. Then \(\sqrt{a^2+b^2}\) is
(a) 7
(b) 13
(c) 5
(d) 25
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Boys \(20\) plus girls with cat eyes \(10\): \(30\) students out of 40.
2. \(\dfrac34\), so \(\sqrt{9+16}=5\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 17: A natural number \(x\) is chosen at random from the first 100 natural numbers. The probability that \(\dfrac{x^2-60x+800}{x-30}<0\) is
(a) \(\dfrac{3}{25}\)
(b) \(\dfrac{7}{25}\)
(c) \(\dfrac{1}{50}\)
(d) \(\dfrac{3}{50}\)
Show Answer & Explanation

Answer: (b) \(\dfrac{7}{25}\)

Explanation:
1. \(x^2-60x+800=(x-20)(x-40)\). The fraction is negative for \(x<20\) and for \(30<x<40\).
2. Values: \(x=1,\dots,19\) (19) and \(x=31,\dots,39\) (9): 28 values.
3. Probability \(\dfrac{28}{100}=\dfrac7{25}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 18: If \(p\) is chosen at random in the closed interval \([0,5]\), the probability that the equation \(x^2+px+\dfrac14(p+2)=0\) has real roots is
(a) \(\dfrac12\)
(b) \(\dfrac14\)
(c) \(\dfrac25\)
(d) \(\dfrac35\)
Show Answer & Explanation

Answer: (d) \(\dfrac35\)

Explanation:
1. Discriminant \(p^2-(p+2)\ge0\), i.e. \((p-2)(p+1)\ge0\).
2. In \([0,5]\): \(p\in[2,5]\), probability \(\dfrac35\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 19: A square is inscribed in a circle. If \(p_1\) is the probability that a randomly chosen point of the circle lies within the square, then \(p_1-p_2\), where \(p_2\) is the probability that the point lies outside the square, equals
(a) \(\dfrac2\pi-1\)
(b) \(\dfrac4\pi-1\)
(c) \(1-\dfrac2\pi\)
(d) \(\dfrac\pi4-1\)
Show Answer & Explanation

Answer: (b) \(\dfrac4\pi-1\)

Explanation:
1. For radius \(r\), the square has area \(2r^2\) and the circle \(\pi r^2\).
2. \(p_1=\dfrac2\pi,\ p_2=1-\dfrac2\pi\), so \(p_1-p_2=\dfrac4\pi-1\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 20: If \(\dfrac{1+4p}4,\ \dfrac{1-p}4,\ \dfrac{1-2p}4\) are probabilities of three mutually exclusive events, the possible values of \(p\) belong to the set
(a) \(\left(0,\dfrac23\right)\)
(b) \(\left[0,\dfrac12\right]\)
(c) \(\left[-\dfrac14,\dfrac12\right]\)
(d) \(\left[-\dfrac23,\dfrac23\right]\)
Show Answer & Explanation

Answer: (c) \(\left[-\dfrac14,\dfrac12\right]\)

Explanation:
1. Each probability lies in \([0,1]\): \(p\ge-\dfrac14\), \(p\le\dfrac34\), \(p\le1\), \(p\le\dfrac12\).
2. The sum \(\dfrac{3+p}4\le1\) gives \(p\le1\). Hence \(-\dfrac14\le p\le\dfrac12\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 21: The probabilities of three events \(A,B,C\) are \(P(A)=0.6,\ P(B)=0.4,\ P(C)=0.5\). If \(P(A\cup B)=0.8,\ P(A\cap C)=0.3,\ P(A\cap B\cap C)=0.2\) and \(P(A\cup B\cup C)\ge0.85\), then the range of \(P(B\cap C)\) is
(a) \(0.2\le P(B\cap C)\le0.25\)
(b) \(0.25\le P(B\cap C)\le0.35\)
(c) \(0\le P(B\cap C)\le0.2\)
(d) \(0.2\le P(B\cap C)\le0.35\)
Show Answer & Explanation

Answer: (d) \(0.2\le P(B\cap C)\le0.35\)

Explanation:
1. \(P(A\cap B)=0.6+0.4-0.8=0.2\).
2. \(P(A\cup B\cup C)=1.5-0.2-0.3-P(B\cap C)+0.2=1.2-P(B\cap C)\ge0.85\) gives \(P(B\cap C)\le0.35\).
3. Also \(P(B\cap C)\ge P(A\cap B\cap C)=0.2\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 22: Ankur and Rahul have an equal number \(n\) of sons. Three mangoes are distributed among all the sons, each mango going to a different son. The probability that two mangoes go to the sons of one friend and one mango to a son of the other is \(\dfrac67\). The number of sons each friend has is
(a) 3
(b) 5
(c) 6
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Total ways \({}^{2n}C_3\). Favourable: choose the friend (2 ways), two of his sons \({}^nC_2\), one of the other's \(n\): \(n^2(n-1)\).
2. Probability \(=\dfrac{3n}{2(2n-1)}=\dfrac67\) gives \(21n=24n-12\), so \(n=4\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 23: A bag contains four tickets numbered 00, 01, 10, 11. A ticket is drawn and replaced; in this way five tickets are drawn. If the probability that the sum of the numbers on the tickets drawn is 23 is \(P\), then the integer part of \(\dfrac1{10P}\) is
(a) 0
(b) 2
(c) 1
(d) 3
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. The numbers are \(0,1,10,11\). Sum 23 in five draws: counting ordered draws gives 100 favourable out of \(4^5=1024\).
2. \(P=\dfrac{25}{256}\), so \(\dfrac1{10P}=\dfrac{256}{250}=1.024\) and its integer part is 1.

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 24: Arti and Bharti are two candidates seeking admission. The probability that Arti is selected is 0.5 and the probability that both are selected is at most 0.3. If the probability of Bharti being selected is \(P\), then the maximum possible value of \(10P\) is
(a) 8
(b) 5
(c) 3
(d) 7
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. \(P(A\cup B)\le1\) gives \(P\le1-0.5+P(A\cap B)\).
2. With \(P(A\cap B)\le0.3\): \(P\le0.8\), so \(10P=8\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 25: Two dice A and B both have six faces. Die A has 3 faces marked 1, 2 faces marked 2 and 1 face marked 3. Die B has 1 face marked 1, 2 faces marked 2 and 3 faces marked 3. Both dice are thrown once; E is the event that the sum equals \(x\). Among \(x=3,4,5,6\), the probability \(P(E)\) is maximum when \(x\) equals
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. \(P(A=1,2,3)=\dfrac12,\dfrac13,\dfrac16\) and \(P(B=1,2,3)=\dfrac16,\dfrac13,\dfrac12\).
2. \(P(3)=\dfrac29,\ P(4)=\dfrac7{18},\ P(5)=\dfrac29,\ P(6)=\dfrac1{12}\); the maximum is at \(x=4\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 26: Two dice A and B both have six faces. Die A has 3 faces marked 1, 2 faces marked 2 and 1 face marked 3. Die B has 1 face marked 1, 2 faces marked 2 and 3 faces marked 3. Both dice are thrown once; E is the event that the sum equals \(x\). Among \(x=3,4,5,6\), the probability \(P(E)\) is minimum when \(x\) equals
(a) 3
(b) 4
(c) 6
(d) 5
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. \(P(3)=\dfrac29,\ P(4)=\dfrac7{18},\ P(5)=\dfrac29,\ P(6)=P(A=3)P(B=3)=\dfrac1{12}\).
2. The minimum among these is \(\dfrac1{12}\) at \(x=6\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 27: The squares of an \(8\times8\) chess board are painted red or black at random. The probability that no column has all its squares alternating in colour is
(a) \(\dfrac1{2^{56}}\)
(b) \(1-\dfrac1{2^7}\)
(c) \(\left(1-\dfrac1{2^7}\right)^8\)
(d) none of these
Show Answer & Explanation

Answer: (c) \(\left(1-\dfrac1{2^7}\right)^8\)

Explanation:
1. A column of 8 squares alternates in \(2\) of \(2^8\) colourings: probability \(\dfrac1{2^7}\).
2. The 8 columns are independent: \(\left(1-\dfrac1{2^7}\right)^8\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 28: A bag contains 4 red and 6 white balls. Two balls are drawn at random. The probability that one of them is red and the other is white is
(a) \(\dfrac{31}{32}\)
(b) \(\dfrac{75}{198}\)
(c) \(\dfrac{8}{15}\)
(d) \(\dfrac{5}{204}\)
Show Answer & Explanation

Answer: (c) \(\dfrac{8}{15}\)

Explanation:
1. \(\dfrac{4\cdot6}{{}^{10}C_2}=\dfrac{24}{45}=\dfrac8{15}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 29: A bag contains 6 apples, 4 bananas and 8 mangoes. If three fruits are drawn at random, the probability that all three are apples is
(a) \(\dfrac{8}{15}\)
(b) \(\dfrac{5}{204}\)
(c) \(\dfrac{31}{32}\)
(d) \(\dfrac{75}{198}\)
Show Answer & Explanation

Answer: (b) \(\dfrac{5}{204}\)

Explanation:
1. \(\dfrac{{}^6C_3}{{}^{18}C_3}=\dfrac{20}{816}=\dfrac5{204}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 30: Five coins are tossed together. The probability that at least one tail will appear is
(a) \(\dfrac{8}{15}\)
(b) \(\dfrac{75}{198}\)
(c) \(\dfrac{5}{204}\)
(d) \(\dfrac{31}{32}\)
Show Answer & Explanation

Answer: (d) \(\dfrac{31}{32}\)

Explanation:
1. \(1-P(\text{no tail})=1-\dfrac1{32}=\dfrac{31}{32}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Question 31: A team of 5 is to be constituted out of 6 girls and 6 boys. The probability that the team has 3 girls and 2 boys is
(a) \(\dfrac{75}{198}\)
(b) \(\dfrac{8}{15}\)
(c) \(\dfrac{31}{32}\)
(d) \(\dfrac{5}{204}\)
Show Answer & Explanation

Answer: (a) \(\dfrac{75}{198}\)

Explanation:
1. \(\dfrac{{}^6C_3\cdot{}^6C_2}{{}^{12}C_5}=\dfrac{20\cdot15}{792}=\dfrac{25}{66}=\dfrac{75}{198}\).

Teacher's Note:
1. List the sample space or count favourable and total cases carefully before dividing.
2. For events, draw a Venn diagram and use inclusion-exclusion to connect the given probabilities.

Free study material for Probability

Probability Objective Questions & Solutions for JEE Mathematics

JEE Mathematics Probability Objective Test Questions

Review structured objective questions for JEE Mathematics Probability. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

NCERT-Aligned Objective Questions and Solutions

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Mathematics to ensure absolute clarity across all sub-topics in this chapter.

Next Steps in Your Exam Preparation

Follow up your worksheet practice by attempting the interactive online Mathematics MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Mathematics Probability MCQs Set 28?

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