GMAT Probability Questions With Solutions, Download PDF

The number of days between 31st January 1972 and 31st January 1977 is 366+365+365+365+366 = 1827

1827 mod 7 = 0

So, 1st February 1972 is a Tuesday and February 1972 has 5 Tuesdays

Required probability = $$\frac{^4 c_3 + ^6 c_3}{^{15} c_3}$$
=(4+20)/(15×14×13)×6=(24×6)/(15×14×13)=24/455 . Choice (b)

Required probability =$$\frac{^6 c_1 * ^5 c_1 + ^5 c_1*^4 c_1 + ^6 c_1*4 c_1 }{^{15} c_2}$$
=$$\frac{30+20+24}{15*7}=74/105$$.
Choice (d)

P (Rahul passes and Manish fails) = 4/9*4/7
P (Rahul fails and Manish passes) = 5/9*3/7
Total probability = 16/63 + 15/63 = 31/63

P (question is solved) = 1 - P (no student solves the question) = 1 - 2/3*3/5*1/2 = 1 - 1/5 = 4/5

There are 3 odd numbers out of total 6 numbers on a die.
P(odd number on both the die) = (3/6)(3/6) = 1/4 = 0.25

Total number of balls = 5+7+8=20
Total number non-red balls = 7+8 = 15
Three balls can be chosen in $$^{20} C _3$$ ways = 1140
Three non-red balls can be chosen in $$^{15} C _3$$ ways = 455
Hence, probability = 455/1140 = 91 / 228
Hence, none of the options is correct.

Total number of balls = 5+7+8=20
Three balls can be chosen in $$^{20} C _3$$ ways = 1140
Two green balls can be chosen in $$^8 C_2$$ ways = 28
One blue ball can be chosen in $$^7 C _1$$ ways = 7
Hence, probability = (28*7) / 1140 = 49/285

Total no. of ways of selecting 52 cards = $$^{52}C_2 = \frac{52\times51}{2} = 1326$$
Total no. of ways of selecting 2 Aces = $$^{4}C_2 = \frac{4\times3}{2} = 6$$
Thus, the required probability = 6/1326 = 1/221

Total number of favorable cases = 6c3*5c1 = 20*5 = 100
Total number of ways in which four balls can be picked = 14c4 = 1001
==> probability = (100/1001) = ~ 0.1

January has 31 days. So it has 3 days more than 4 weeks.
For a January to have 5 Wednesdays the 1st day must be either a Monday or a Tuesday or a Wednesday.
Thus the probability = 3/7.

No. of ways of drawing 2 socks = $$^{20}C_2$$
No. of pairs of socks = 10
=> Probability = $$\frac{10}{^{20}C_2} = \frac{1}{19}$$

Since both the events are mutually exclusive, the two probabilities need to be multiplied for calculating the probability of a member being female-engineer.
=> P(female-engg) = P(engg)XP(female) = 0.45 X 0.4 = 0.18

If we remove 2 books from k, there will be k-2 books left.
Now there will be k-2+1 = k - 1 spaces which will be created. None of these will be adjacent to each other. Hence, we can place 2 books in any two of these k -1 spaces. Hence the number of ways of doing this would be (k-1) C 2
Total ways of selecting 2 books from ‘k’ books = k C 2.
We have been given that
$$\frac{(k-1) C 2}{k C 2} = \frac{5}{7}$$
=> $$\frac{(k-1)*(k-2)}{k*(k-1)} = \frac{5}{7}$$
=> $$\frac{k-2}{k} = \frac{5}{7}$$
=> 7k - 14 = 5k
=> 2k = 14
=> k = 7
Hence there are 7 books in the shelf.

Here, the maximum possible outcome is 18.
The number of cases when the sum of three outcomes is 18 = 1
The number of cases when the sum of three outcomes is 17 = $$^3c_1 $$ (when the outcome of two attempts is 6 and one is 5).
Number of possible outcome = 6*6*6 = 216
Required probability = $$\frac{216-4}{216}$$ = $$\frac{53}{54}$$

Probability of getting a head = $$\frac{2}{3}$$
Probability of getting a tail = $$\frac{1}{3}$$
Number of ways of getting two heads in three attempts = 3C2 = 3
Probability of getting exactly two heads = $$\frac{2}{3}* \frac{2}{3} *\frac{1}{3} *3=\frac{4}{9}$$
Hence Option D.

Out of the 16 eggs, 5 eggs are rotten and 11 eggs are in good condition.

According to the question, out of the two eggs drawn one is rotten and the other is in good condition.

Hence, required probability = $$ \frac{^5C_{1} * ^{11}C_{1}}{^{16}C_{2}} = \frac{5*11}{16*15/2} = \frac{11}{24}$$

Hence, option A is the right choice.

Probability of Ram winning 1 game at most = Probability of Ram losing all the 3 games + Probability of Ram winning exactly 1 game.
The only even prime is 2.
Probability of winning a game = 1/6.
Probability of losing a game = 5/6.

Ram can lose all the 3 games in only 1 way.
Hence, probability of losing = 1*(5/6)*(5/6)*(5/6) = 125/216.

The number of ways in which exactly 1 game can be won = 3 (First game or second game or third game can be won by Ram).
Therefore, probability of winning exactly 1 game = 3*(1/6)(5/6)(5/6) = 75/216

Total probability of winning at most 1 game = 125/216 + 75/216 = 200/216 =25/27.

Number of ways of drawing 4 marbles out of 16

=> $$n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$$

= $$1820$$

Out of the four drawn marbles, 2 are red and 2 are blue.

=> $$n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$$

= $$28 \times 3 = 84$$

$$\therefore$$ Required probability = $$\frac{n(E)}{n(S)}$$

= $$\frac{84}{1820} = \frac{3}{65}$$

Number of ways of drawing 2 marbles out of 16

$$n(S) = C^{16}_2 = \frac{16 \times 15}{1 \times 2}$$

= $$120$$

Out of the two drawn marbles, both are green

=> $$n(E) = C^5_2 = \frac{5 \times 4}{1 \times 2}$$

= $$10$$

$$\therefore$$ Required probability = $$\frac{n(E)}{n(S)}$$

= $$\frac{10}{120} = \frac{1}{12}$$

Number of ways of drawing 3 marbles out of 16

$$n(S) = C^{16}_3 = \frac{16 \times 15 \times 14}{1 \times 2 \times 3}$$

= $$560$$

Out of the three drawn marbles, none is red, i.e., they will be either blue or green.

=> $$n(E) = C^8_3 = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}$$

= $$56$$

$$\therefore$$ Required probability = $$\frac{n(E)}{n(S)}$$

= $$\frac{56}{560} = \frac{1}{10}$$

A number can end with {0, 1, 2, 3 ... 9}

If all three number ends with the same digit then the unit digit of their product will be same as the product of the unit digit of all three numbers.

If unit digit is 0 then unit digit of the product = 0*0*0 = 0

If unit digit is 1 then unit digit of the product = 1*1*1 = 1

If unit digit is 2 then unit digit of the product = 2*2*2 = 8

If unit digit is 3 then unit digit of the product = 3*3*3 = 27

If unit digit is 4 then unit digit of the product = 4*4*4 = 64

If unit digit is 5 then unit digit of the product = 5*5*5 = 125

If unit digit is 6 then unit digit of the product = 6*6*6 = 216

If unit digit is 7 then unit digit of the product = 7*7*7 = 343

If unit digit is 8 then unit digit of the product = 8*8*8 = 512

If unit digit is 9 then unit digit of the product = 9*9*9 = 729

We can see that there are 6 favourable outcomes. {0, 1, 4, 5, 6, 9}

Therefore the probability = $$\frac{6}{10}$$ = $$\frac{3}{5}$$

Hence we can say that option D is the correct answer.

This question is asking about at least one person will pass the exam. It means if we exclude or subtract those cases which are saying will pass the exam. Then we will get it.
So at least one person will pass the exam = 1 - None of them pass the exam ….Eq(1)
None of them pass the exam = A will not pass $$\times$$ B will not pass $$\times$$ B will not pass
= (1-$$\frac{1}{2}$$) $$\times$$ (1-$$\frac{1}{3}$$) $$\times$$ (1-$$\frac{1}{5}$$)
= $$\frac{1}{2}\times\frac{2}{3}\times\frac{4}{5}$$
= $$\frac{4}{15}$$ ….Eq(2)
Put Eq(2) in Eq(1).
So at least one person will pass the exam = 1 - None of them pass the exam
= 1 - $$\frac{4}{15}$$
= $$\frac{15-4}{15}$$
= $$\frac{11}{15}$$
Hence, option b is the correct answer.

Probability of picking 1 blue ball from ‘x’ green balls and ‘x+2’ blue balls = $$\dfrac{x+2}{x+x+2} = \dfrac{x+2}{2x+2}$$

Given, $$\dfrac{x+2}{2x+2} = \dfrac{5}{9}$$

$$9x+18 = 10x+10$$
$$x = 8$$.

Therefore, Probability of picking x-7 blue balls = Probability of picking 8-7 = 1 blue ball = $$\dfrac{5}{9}$$.

Let P1 be the probability of a student getting out from the Blue house = 10/30 = ⅓
Let P2 be the probability of a student getting out from the Orange house = 5/30 = ⅙
So the probability of a student getting out either from the Blue or Orange house = P1 or P2
(P1 or P2) = P1 + P2
= ⅓ + ⅙
= ½

If there are to be at least 8 females remaining in the class, then the possibilities can be either 8, 9, or 10 females in the class, including the teacher. If 3 students are out of class then the possibilities will be (any 2 are females and 1 student is male are out of class) or (any 1 student is female and 2 are males are out of the class) or (all 3 are male students are out of the class) and the total number of persons in the class including a teacher will be 30 + 1 = 31.

$$\dfrac{9_{C_2}\times21_{C_1}+9_{C_1}\times21_{C_2}+21_{C_3}}{^{30}C_3}\ =\ \dfrac{756\ +\ 1890\ +\ 1330}{4060}\ =\ \dfrac{3976}{4060}\ =\ \dfrac{142}{145}$$

The probability that at least 8 females (including the teacher) remain in the class is 142/145.

Probability of failing the exam by A = $$1-\frac{1}{4}$$ = $$\frac{3}{4}$$

Probability of failing the exam by B = $$1-\frac{3}{5}$$ = $$\frac{2}{5}$$

Probability of failing the exam by C = $$1-\frac{1}{6}$$ = $$\frac{5}{6}$$

Here at least two people fail in the examination means any one person among three pass the exam or all people fail in the exam.

Probability that at least two people fail in the examination = $$\frac{1}{4}\times\frac{2}{5}\times\frac{5}{6}+\frac{3}{4}\times\frac{3}{5}\times\frac{5}{6}+\frac{3}{4}\times\frac{2}{5}\times\frac{1}{6}+\frac{3}{4}\times\frac{2}{5}\times\frac{5}{6}$$

= $$\frac{10}{120} + \frac{45}{120} + \frac{6}{120} + \frac{30}{120}$$

= $$\frac{91}{120}$$

Hence, option d is the correct answer.

Given, There are T black balls, T-7 blue balls and 12 red balls.
Probability of taking 1 blue and 1 red ball = $$\dfrac{(T-7)(12)}{(T+T-7+12)_{C_{2}}}$$

$$\dfrac{12T-84}{2T+5_{C_{2}}} = \dfrac{(12T-84)\times2}{(2T+5)\times(2T+4)}$$

$$= \dfrac{24T-168}{4T^2+18T+20} = \dfrac{6}{37}$$

$$ ⇒ 37(24T-168) = 6(4T^2+18T+20)$$

$$ ⇒ 888T - 6216 = 24T^2+108T+120$$

$$ ⇒ 24T^2-780T+6336 = 0$$

Solving the above equation, we get T = 16.5 or 16.

Since the number of balls cannot be in decimals, the value of T = 16.

If a person wants Rs.13 and two five-rupee coins should be there, then there will be two possibilities.
Possibility I: Two 5-rupee coins and 3 one-rupee coins
$$\dfrac{3_{C_{2}}\times5_{C_{3}}}{18_{C_{5}}} = \dfrac{3\times10}{8568} = \dfrac{5}{1428}$$

Possibility II: Two 5-rupee coins, 1 one-rupee coin and 1 two-rupee coin.
$$\dfrac{3_{C_{2}}\times5_{C_{1}}\times10_{C_{1}}}{18_{C_{4}}} = \dfrac{3\times5\times10}{3060} = \dfrac{5}{102}$$

Hence, The required probability = $$\dfrac{5}{1428}+\dfrac{5}{102} = \dfrac{70+5}{1428} = \dfrac{75}{1428} = \dfrac{25}{476}$$.

Probability of picking three balls from the bag such that none of them is red = $$\frac{17_{C_{_{3}}}}{21_{C_{_{3}}}}$$

= $$\frac{17\times16\times15}{21\times20\times19}$$

= $$\frac{68}{133}$$
Hence, option b is the correct answer.

Number of blue balls = 7
Number of red balls = 5
Number of green balls = 8
Probability of taking 3 balls such that 2 are green = $$\dfrac{8_{C_{2}}\times12}{20_{C_{3}}}$$
$$\dfrac{28\times12}{1140} = \dfrac{28}{95}$$

The number of prime numbers less than 100 are 25. Of these 6 of them - 7, 17, 37, 47, 67 and 97 have 7 as last digit. Hence, p = 6 / 25

Number of red balls = 4.
Number of green balls = 5.
Number of blue balls = 3.
Probability of taking two different coloured balls = $$\dfrac{4\times5+5\times3+4\times3}{12_{C_{2}}}$$

$$\dfrac{20+15+12}{66} = \dfrac{47}{66}$$.

Probability = $$\frac{9_{C_{_{1}}} \times 11_{C_{_{1}}} + 11_{C_{_{2}}} }{20_{C_{_2}}}$$

= $$\frac{9\times11 + 55 }{190}$$

= $$\frac{99 + 55 }{190}$$

= $$\frac{154 }{190}$$

= $$\frac{77}{95}$$
Hence, option d is the correct answer.

The probability of winning the lottery by Neha is $$\frac{2}{9}$$.
The probability of losing the lottery by Neha = 1 - $$\frac{2}{9}$$ = $$\frac{7}{9}$$
The probability of losing the lottery by Nitu is $$\frac{5}{6}$$.
The probability of winning the lottery by Nitu = 1 - $$\frac{5}{6}$$ = $$\frac{1}{6}$$

Probability that at least one of them will win the lottery = $$\frac{2}{9}\times\frac{1}{6} + \frac{7}{9}\times\frac{1}{6} + \frac{2}{9}\times\frac{5}{6}$$

= $$\frac{2}{54} + \frac{7}{54} + \frac{10}{54} $$

= $$\frac{19}{54} $$

Hence, option c is the correct answer.

If the probability of passing the exam by Rina is $$\frac{2}{13}$$.
Probability of failing the exam by Rina = $$1 - \frac{2}{13}$$ = $$\frac{11}{13}$$
The probability of failing the exam by Mina is $$\frac{6}{11}$$.
probability of passing the exam by Mina = $$1 - \frac{6}{11}$$ = $$\frac{5}{11}$$
Probability that at least one of them will pass in the exam = $$\frac{2}{13} \times \frac{6}{11} + \frac{11}{13} \times \frac{5}{11} + \frac{2}{13} \times \frac{5}{11}$$

= $$\frac{12+55+10}{143}$$

= $$\frac{77}{143}$$
Hence, option c is the correct answer.

Total number of balls = 10 + 7 + 3 = 20
When two balls are picked at random,
Probability of not getting yellow ball = $$\dfrac{^{13}C_2}{^{20}C_2}$$
Probability of getting atleast one yellow ball = 1 - Probability of not getting yellow ball
= 1 - $$\dfrac{^{13}C_2}{^{20}C_2}$$
= 1 - $$\dfrac{13\times12}{20\times19}$$
= 1 - $$\dfrac{39}{95}$$
= $$\dfrac{56}{95}$$
Hence, the correct answer is Option E

Total number of pencils = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 pencils out of 7
= 7C2 `

= $$\ \dfrac{\ 7\times\ 6}{2\times\ 1}$$

= 21.
Let E = Event of drawing 2 pencils, none of which is yellow.

n(E) = Number of ways of drawing 2 pencils out of (2 + 3) pencils
= 5C2

= $$\ \dfrac{\ 5\times\ 4}{2\times\ 1}$$

= 10.
P(E) = $$\ \dfrac{\ n\left(E\right)}{n\left(S\right)}$$ = $$\ \ \dfrac{\ 10}{21}$$

Hence answer is option b

If sum should be more than 9. It means that it can be 10, 11 and 12.
10 ⇒ (6,4) (5,5) (4,6)
11 ⇒ (6,5) (5,6)
12 ⇒ (6,6)

Probability = $$\frac{6}{36}$$
= $$\frac{1}{6}$$
Hence, option d is the correct answer.

Total number of balls = 6 + 8 + 9 + 7 = 30
Total number of outcomes = $$^{30}$$C$$_4$$
Number of favourable outcomes = $$^6$$C$$_4$$ + $$^8$$C$$_4$$ + $$^9$$C$$_4$$ + $$^7$$C$$_4$$
Required probability = $$\frac{^6C_4 + ^8C_4 + ^9C_4+ ^7C_4}{^{30}C_4}$$
= $$\frac{15 + 70 + 126 + 35}{27405}$$
= $$\frac{246}{27405}$$ = $$\frac{82}{9135}$$
Hence, the correct answer is Option C

Probability = $$\dfrac{(y+6)(y+5)(y+4)}{(2y+6)(2y+5)(2y+4)} = \dfrac{204}{1015}$$

$$-617y^{3}+2985y^{2}+44918y+97320 = 0$$

After solving the above equation, y = 12.
Hence, option c is the correct answer.

In a bag, there a total of 36 bottles of four different colors.
The number of red bottles is 10.
red+blue+yellow+green = 36
10+blue+yellow+green = 36
blue+yellow+green = 36-10 = 26
Probability of picking two bottles from the bag such as alteast one of them is red = $$\frac{10_C{_{_1}}\times26_C{_{_1}}+10_C{_{_2}}}{36_C{_{_2}}}$$
= $$\frac{10\times26+\frac{10\times9}{2\times1}}{\frac{36\times35}{2\times1}}$$
= $$\frac{10\times26+5\times9}{18\times35}$$
= $$\frac{260+45}{630}$$
= $$\frac{305}{630}$$
= $$\frac{61}{126}$$
Hence, option c is the correct answer.