{"id":26454,"date":"2019-03-25T16:26:01","date_gmt":"2019-03-25T10:56:01","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26454"},"modified":"2019-03-25T16:26:01","modified_gmt":"2019-03-25T10:56:01","slug":"ssc-cgl-dice-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/ssc-cgl-dice-questions-pdf\/","title":{"rendered":"SSC CGL Dice Questions PDF"},"content":{"rendered":"

SSC CGL Dice Questions PDF:<\/strong><\/span><\/h1>\n

Download SSC CGL Dice questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Dice objective questions for SSC exams.<\/p>\n

Download SSC CGL Dice Questions PDF<\/a><\/p>\n

Take a free SSC CGL mock test<\/a><\/p>\n\n

Question 1:\u00a0<\/b>A Dice is thrown. Find the probability that the number showing on the dice is divisible by 2<\/p>\n

a)\u00a01\/4<\/p>\n

b)\u00a01\/6<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a01\/3<\/p>\n

e)\u00a0None of these<\/p>\n

Question 2:\u00a0<\/b>In a single throw of two dice, find the probability of getting doublet?<\/p>\n

a)\u00a02\/6<\/p>\n

b)\u00a03\/5<\/p>\n

c)\u00a06\/7<\/p>\n

d)\u00a01\/6<\/p>\n

e)\u00a0None of these<\/p>\n

Question 3:\u00a0<\/b>Two dice are tossed. Find the probability that the total is a prime number?<\/p>\n

a)\u00a05\/12<\/p>\n

b)\u00a07\/12<\/p>\n

c)\u00a012\/16<\/p>\n

d)\u00a06\/16<\/p>\n

e)\u00a0None of these<\/p>\n

Question 4:\u00a0<\/b>In simultaneous throw of a pair of a dice, find the probability that the sum of numbers shown on the two faces divisible by 5 or 6<\/p>\n

a)\u00a013\/36<\/p>\n

b)\u00a01\/2<\/p>\n

c)\u00a01\/6<\/p>\n

d)\u00a01\/36<\/p>\n

e)\u00a0None of these<\/p>\n

Question 5:\u00a0<\/b>4 dice are thrown and the sum of the numbers noted is 10. Find the probability that all the numbers lie between 2 and 5 (both inclusive)?<\/p>\n

a)\u00a00.275<\/p>\n

b)\u00a00.175<\/p>\n

c)\u00a00.250<\/p>\n

d)\u00a00.125<\/p>\n

SSC CGL Previous Papers Download PDF<\/a><\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n\n

Question 6:\u00a0<\/b>Find the probability of throwing a sum of less than 8 or at least 11 using 3 different dice.<\/p>\n

a)\u00a0$\\frac{143}{216}$<\/p>\n

b)\u00a0$\\frac{123}{216}$<\/p>\n

c)\u00a0$\\frac{73}{216}$<\/p>\n

d)\u00a0$\\frac{93}{216}$<\/p>\n

Question 7:\u00a0<\/b>Three dice are rolled. It is known that the sum obtained is 14. What is the probability that the first die shows a 5?<\/p>\n

a)\u00a04\/15<\/p>\n

b)\u00a01\/3<\/p>\n

c)\u00a01\/5<\/p>\n

d)\u00a02\/5<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 8:\u00a0<\/b>2 dice are cast. What is the probability that both the dice show the same number?<\/p>\n

a)\u00a01\/2<\/p>\n

b)\u00a02\/3<\/p>\n

c)\u00a01\/3<\/p>\n

d)\u00a01\/6<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 9:\u00a0<\/b>4 dice are rolled. In how many ways can a sum of 22 be obtained?<\/p>\n

a)\u00a04<\/p>\n

b)\u00a02<\/p>\n

c)\u00a026<\/p>\n

d)\u00a016<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 10:\u00a0<\/b>Two persons A and B play a game of dice in which each person throws a die on its turn and the person who throws 6 first wins the game. A starts the game. Find the probability that A wins.<\/p>\n

a)\u00a05\/11<\/p>\n

b)\u00a06\/11<\/p>\n

c)\u00a07\/11<\/p>\n

d)\u00a04\/11<\/p>\n

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Question 11:\u00a0<\/b>A fair dice containing six sides is thrown two times. What is the probability that the sum of the two rolls of dice is equal to 2?<\/p>\n

a)\u00a0$\\frac{1}{6}$<\/p>\n

b)\u00a0$\\frac{1}{36}$<\/p>\n

c)\u00a0$\\frac{1}{18}$<\/p>\n

d)\u00a0$\\frac{1}{9}$<\/p>\n

e)\u00a0$\\frac{1}{12}$<\/p>\n

Question 12:\u00a0<\/b>Two dice are thrown and the numbers that appear on them are a and b respectively. Find the probability that the sum of a and b is 8.<\/p>\n

a)\u00a0$\\frac{5}{6}$<\/p>\n

b)\u00a0$\\frac{1}{6}$<\/p>\n

c)\u00a0$\\frac{5}{36}$<\/p>\n

d)\u00a0$\\frac{31}{36}$<\/p>\n

Question 13:\u00a0<\/b>When two dice are rolled, what is the probability that the sum of numbers that appear is greater than 8?<\/p>\n

a)\u00a0$\\frac{13}{18}$<\/p>\n

b)\u00a0$\\frac{11}{18}$<\/p>\n

c)\u00a0$\\frac{5}{18}$<\/p>\n

d)\u00a0$\\frac{7}{18}$<\/p>\n

Question 14:\u00a0<\/b>Find the smallest positive integer that should be multiplied to 2352 to make it a perfect square.<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a07<\/p>\n

d)\u00a011<\/p>\n

Question 15:\u00a0<\/b>When three dice are rolled, find the probability that the sum of numbers that appear on them is equal to 10.<\/p>\n

a)\u00a01\/4<\/p>\n

b)\u00a01\/8<\/p>\n

c)\u00a01\/12<\/p>\n

d)\u00a01\/16<\/p>\n

SSC CGL Free Mock Test<\/a><\/p>\n\n

Question 16:\u00a0<\/b>If three distinct fair dice are rolled, in how many ways can a sum of 16 be obtained?<\/p>\n

a)\u00a03 ways<\/p>\n

b)\u00a09 ways<\/p>\n

c)\u00a018 ways<\/p>\n

d)\u00a06 ways<\/p>\n

Question 17:\u00a0<\/b>A man throws a dice and picks a card from a pack of 52 cards randomly. What is the probability that he gets an even number from the dice and a King?<\/p>\n

a)\u00a01\/13<\/p>\n

b)\u00a01\/26<\/p>\n

c)\u00a01\/39<\/p>\n

d)\u00a01\/52<\/p>\n

e)\u00a0none of these<\/p>\n

Question 18:\u00a0<\/b>What is the probability of getting an odd number when a dice is thrown twice?<\/p>\n

a)\u00a01\/4<\/p>\n

b)\u00a01\/3<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a03\/4<\/p>\n

e)\u00a0none of these<\/p>\n

Question 19:\u00a0<\/b>When a dice is thrown twice, what is the probability that the sum of the numbers obtained is 6 ?<\/p>\n

a)\u00a01\/12<\/p>\n

b)\u00a01\/18<\/p>\n

c)\u00a01\/36<\/p>\n

d)\u00a01\/9<\/p>\n

e)\u00a05\/36<\/p>\n

Question 20:\u00a0<\/b>Two dices are thrown simultaneously, what is the probability that the sum of the numbers that turn up is less than 9?<\/p>\n

a)\u00a0$\\frac{5}{18}$<\/p>\n

b)\u00a0$\\frac{2}{3}$<\/p>\n

c)\u00a0$\\frac{25}{36}$<\/p>\n

d)\u00a0$\\frac{13}{18}$<\/p>\n

100+ Free GK Tests for SSC Exams<\/a><\/p>\n

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Question 21:\u00a0<\/b>A dice is rolled twice, what is the probability of getting 3 at least once?<\/p>\n

a)\u00a025\/36<\/p>\n

b)\u00a011\/36<\/p>\n

c)\u00a013\/36<\/p>\n

d)\u00a023\/36<\/p>\n

e)\u00a0None of these<\/p>\n

Question 22:\u00a0<\/b>A dice is thrown twice. What is the probability that sum of the numbers which turn up is 6.<\/p>\n

a)\u00a07\/36<\/p>\n

b)\u00a05\/36<\/p>\n

c)\u00a01\/4<\/p>\n

d)\u00a01\/9<\/p>\n

e)\u00a0none of these<\/p>\n

Question 23:\u00a0<\/b>Two dice are rolled simultaneously. What is the probability that none of the dice shows up a four?<\/p>\n

a)\u00a0$\\frac{5}{6}$<\/p>\n

b)\u00a0$\\frac{11}{36}$<\/p>\n

c)\u00a0$\\frac{1}{2}$<\/p>\n

d)\u00a0$\\frac{25}{36}$<\/p>\n

Question 24:\u00a0<\/b>Two fair dice are rolled simultaneously. What is the probability that atleast one of the dice shows up a three?<\/p>\n

a)\u00a0$\\frac{5}{6}$<\/p>\n

b)\u00a0$\\frac{11}{36}$<\/p>\n

c)\u00a0$\\frac{1}{2}$<\/p>\n

d)\u00a0$\\frac{25}{36}$<\/p>\n

Question 25:\u00a0<\/b>When two dice are thrown simultaneously what is the probability that the product obtained by multiplying the outcomes is an odd number?<\/p>\n

a)\u00a02\/5<\/p>\n

b)\u00a03\/4<\/p>\n

c)\u00a01\/2<\/p>\n

d)\u00a01\/4<\/p>\n

e)\u00a0none of these<\/p>\n

1500+ Free SSC Questions & Answers<\/a><\/p>\n\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Sample spaces = {1, 2, 3, 4, 5, 6} n(S) = 6
\nNumbers divisible by 2 are 2, 4 and 6
\nn(E) =3
\nP(E) = 3\/6=1\/2<\/p>\n

2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Sample spaces = {(1, 1), (1, 2) \u2026 (1, 6)
\n(2, 1), (2, 2) … (2, 6)
\n(3, 1), (3, 2) … (3, 6)
\n(4, 1), (4, 2) … (4, 6)
\n(5, 1), (5, 2) … (5, 6)
\n(6, 1), (6, 2) … (6, 6)}
\nn(S) = 36
\nFavorable cases to get doublets are = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
\nn(E) = 6
\nProbability = n(E)\/n(S) =6\/36=1\/6
\n3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

n(s)=((1,1),(1,2),\u2026\u2026.(6,6)) n(s)=36
\nFavorable cases to get prime numbers as sum= {(1,1) (1,2) (1,4) (1,6) (2,1), (2,3) (2,5) (3,2), (3,4) (4,1) (4,3) (5,2) (5,6) (6,1), (6,5)}
\nn(A)= 15 P(A)=n(A)\/n(s) =15\/36=5\/12<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

n(S) = 6\u00d76=36
\nEvent of getting a sum of numbers shown on the two faces divisible by 5 or 6 = [(1,4), (1,5), (2,3), (2,4), (3,2), (3,3), (4,1), (4,2),(4,6),(5,1), (5,5),(6,4), (6,6)]
\nn(E) =13 P(E) = n(E)\/n(S) =13\/36<\/p>\n

5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the numbers thrown on 4 dice be a,b,c,d
\na+b+c+d=10 a,b,c,d lie between 2 and 5.
\nOnly case valid are {2,2,2,4},{2,2,3,3}
\nTotal no. of cases in{2,2,2,4}=4
\nTotal No. of cases in{2,2,3,3}=6
\nTotal cases=10 Now, x+y+z+w=10
\nWhere x,y,z,w lie between 1 and 6.
\nEach of x,y,z,w has to be at least 1.
\nSo the valid cases of the above equation is $^9C_3$
\nBut it will include 4 cases 0f {7,1,1,1}
\nTotal cases= $^9C_3$-4=80
\nHence probability=10\/80=1\/8=0.125<\/p>\n\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

P(X<8 || X>=11) = 1 – P(8) – P(9) – P(10)
\nP(8) => a+b+c = 8 and each of them must be at least 1 and not more than 6 => $^7C_2$ ways = 21 ways
\nP(9) => a+b+c = 9 and each of them must be at least 1 and not more than 6 => ($^8C_2-(3*^2C_2$) ways = 25 ways
\nP(10) => a+b+c = 10 and each of them must be at least 1 and not more than 6 => $(^9C_2 – (3*^3C_2))$ ways = 27 ways
\nTotal possibilities = 21+25+27 = 73
\nP(X<8 || X>=11) = $1-\\frac{73}{216}$ = $\\frac{143}{216}$<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

The possibilities for getting a sum of 14 are:
\n6, 6, 2 -> 3 arrangements possible
\n6, 5, 3 -> 6 arrangements possible
\n6, 4, 4 -> 3 arrangements possible
\n5, 5, 4 -> 3 arrangements possible<\/p>\n

Starting with 5, for the set 6, 5, 3, there are 2 arrangements possible and for the set 5, 5, 4, there are 2 arrangements possible.
\nSo, required probability = (2+2)\/15 = 4\/15<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Number of favourable cases = 6
\nTotal number of cases = 6*6 = 36
\nSo, probability = 1\/6<\/p>\n

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

There should be at least 2 6s. The other 9 can be obtained in 6+3, 5+4 = 2 ways
\nSo, the combinations are 6+6+6+3, which can be arranged in 4!\/3! = 4 ways
\n6+6+5+4, which can be arranged in 4!\/2! = 12 ways.
\nSo, the total number of ways = 4 + 12 = 16 ways<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

P = P(A wins on 1st throw)+P(A wins on 3rd throw)+P(A wins on 5th throw)+…..and so on
\nP = (1\/6)+(5\/6*5\/6*1\/6)+…… = (1\/6)\/[1-(25\/36)] = 6\/11<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The number of different permutations possible with two throws of dice is equal to 6*6=36
\nFor the sum of the rolls to equal 2, both the rolls should be equal to 1.
\nHence, the probability equals $\\frac{1}{36}$<\/p>\n

Free SSC Daily Practice Set<\/a><\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Number of ways such that sum is 8 => (2,6), (3,5), (4,4), (5,3), (6,2)
\n=> 5 possible cases.
\nTotal number of cases = 6*6 = 36
\nProbability = $\\frac{5}{36}$<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Sum is greater than 8 => 9, 10, 11, 12 are possibilities.
\nSum = 9 => (3,6), (6,3), (4,5), (5,4)
\nSum = 10 => (4,6), (6,4), (5,5)
\nSum = 11 => (5,6), (6,5)
\nSum = 12 => (6,6)
\n=> 10 possibilities
\n=> probability = $\\frac{10}{36}$ = $\\frac{5}{18}$<\/p>\n

14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let\u2019s factorize 2352 into its prime factors.
\n2352 = 16 * 3 * 49 = $2^4 * 3 * 7^2 $.
\nTo be a perfect square, the number should have prime factors with even indices. Hence, for 3 to have an even index, the number should be multiplied with 3.<\/p>\n

15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let a, b and c be the numbers that appear on the 3 dice.
\na + b + c = 10
\nEach of a, b and c are at least 1.
\n=> So, the equation changes to x + y + z = 7
\nThis can be done in $^9C_2$ = 36 ways.<\/p>\n

But, none of a, b or c can be more than 6. So, we need to remove cases where any of x, y or z is more than 5.
\nThese cases are:
\nIf (x, y, z) is (7, 0, 0). This can be done in $^3C_2$ = 3 ways.
\nIf (x, y, z) is (6, 1, 0). This can be done in 3! ways = 6 ways.
\nTotal number of ways = 9 ways<\/p>\n

Number of required ways = 36 – 9 = 27<\/p>\n

Total number of combinations = 6 * 6 * 6 = 216<\/p>\n

=> Probability = $\\frac{27}{216}$ = $\\frac{1}{8}$<\/p>\n\n

16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

A sum of 16 can be obtained in the following ways (the maximum on each dice is 6):<\/p>\n

6 + 5 + 5
\n6 + 6 + 4<\/p>\n

The first combination can be arranged among the three dice in 3!\/2! = 3 ways
\nThe second combination can be arranged in 3!\/2! = 3 ways
\nSo, total number of ways in which a sum of 16 can be obtained = 3 + 3 = 6 ways<\/p>\n

17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The probability of getting an even number from the dice = no. of favorable outcomes(2,4,6)\/ total number of outcomes(1,2,3,4,5,6) = P(e) = 3\/6 = 1\/2
\nThe probability of getting a King from a pack of 52 cards = P(k) = no. of favorable outcomes (4 Kings)\/ total number of outcomes(52 cards) = 4\/52 = 1\/13
\nSince, both events are independent, the required probability = P(e) x P(k) = 1\/2 x 1\/13 = 1\/26<\/p>\n

18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let P(O1) be the probability of getting an odd no. when the dice is thrown first and P(O2) when the dice is thrown for the second time.
\nWe need to find the probability that either the first or second time the outcome would be an odd number i.e. P(O1 U O2)
\nSince P(O1) and P(O2) are independent, P(O1 U O2) = P(O1) + P(O2) – P(O1)P(O2) = 3\/6 + 3\/6 – 3\/6 x 3\/6 = 3\/4<\/p>\n

19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

favorable outcomes = E = (1,5) (5,1) (2,4) (4,2) (3,3)
\nThus, no of favorable outcomes = n(E) = 5
\nTotal no. of outcomes = n(S) = 6 x 6 = 36
\nThus, the required probability = n(E)\/n(S) = 5\/36<\/p>\n

20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

When two dices are thrown simultaneously, the total possible outcomes are 6*6 = 36
\nInstead of calculating the outcomes where the sum is less than 9, we can easily calculate the cases where the sum is greater than or equal to 9. Such cases are
\n(5,4), (4,5), (6,3), (3,6), (5,5), (6,4), (4,6), (6,5), (5,6), (6,6)
\nSo there are 10 cases. Hence the probability that sum is greater than or equal to 9 is $\\frac{10}{36}$ = $\\frac{5}{18}$
\nSo the probability that sum is less than 9 is 1- $\\frac{5}{18}$ = $\\frac{13}{18}$<\/p>\n\n

21)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The probability of getting 3 at least once = 1 – probability of not getting 3 in any of the throws.
\n= 1 – $\\frac{5}{6}*\\frac{5}{6}$
\n= 1- 25\/36 = 11\/36
\nSo the correct option to choose is B- 11\/36<\/p>\n

22)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Total possible cases = 6*6 =36
\nFavorable cases = (1,5), (2,4),(3,3),(4,2),(5,1)
\nSo the favorable cases are 5.
\nHence the required probability is 5\/36<\/p>\n

23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Each of the dice has to show up a number other than 4.
\nSo there are five possible outcomes for each of the dice.
\nSo number of favourable ways = 5*5 = 25.
\nTotal number of ways = 6*6 = 36.
\nProbability = 25\/36.<\/p>\n

24)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

We can use the 6×6 matrix to find out the number of ways in which atleast one of the dice shows up a three.
\nThe favourable cells are as highlighted in green.<\/p>\n

<\/p>\n

Thus the number of favourable ways = 11.
\nProbability = 11\/36.<\/p>\n

25)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

For the product to be an odd number, the outcomes of both the dice should be odd.
\nThe favorable Event = E = {(1,1), (1,3), (1,5), (3, 5), (5,3), (5,1), (3,1), (3,3), (5,5) }
\nTotal number of possible outcomes = 6 x 6 = 36
\nThus, the required probability = 9\/36 = 1\/4<\/p>\n

SSC Free Previous Papers App<\/a><\/p>\n\n","protected":false},"excerpt":{"rendered":"

SSC CGL Dice Questions PDF: Download SSC CGL Dice questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Dice objective questions for SSC exams. Question 1:\u00a0A Dice is thrown. Find the probability that the number showing on the dice is divisible by 2 a)\u00a01\/4 b)\u00a01\/6 c)\u00a01\/2 d)\u00a01\/3 […]<\/p>\n","protected":false},"author":21,"featured_media":26476,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,504],"tags":[462,1519,1659,1663],"class_list":{"0":"post-26454","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-cgl","10":"tag-ssc-cgl","11":"tag-ssc-cgl-2019","12":"tag-ssc-cgl-free-mock","13":"tag-ssc-cgl-previous-papers"},"better_featured_image":{"id":26476,"alt_text":"SSC CGL Dice Questions PDF","caption":"SSC CGL Dice Questions PDF","description":"SSC CGL Dice Questions 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