Two dice are thrown. Find the probability of getting an odd number on the first die and a multiple of 3 on the other. -Maths 9th

1 Answer

Answer :

Let A : Getting an odd number on first die; B : Getting a multiple of 3 on second die, thenA = {1, 3, 5}, B = {3, 6}          ∴ P(A) = \(rac{3}{6}=rac{1}{2}\), P(B) = \(rac{2}{6}=rac{1}{3}\)The events A and B are independent∴ Required probability = P (A) . P (B) = \(rac{1}{2}\) x \(rac{1}{3}\) = \(rac{1}{6}\)

Related questions

Description : Two unbiased dice are rolled. Find the probability of getting a multiple of 2 on one die and a multiple of 3 on the other die ? -Maths 9th

Last Answer : When two unbiased dice are rolled, the possible out comes are∴ n(S) = 36 Let A : getting a multiple of 2 on one die and a multiple of 3 on the other die. ⇒ A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), ( ... (3, 6), (6, 2), (6, 4)} ⇒ n(A) = 11∴ P(A) = \(rac{n(A)}{n(S)} =rac{11}{36}.\)

Description : A die is thrown.What is the probability of getting a multiple of 3 on the upper face ? -Maths 9th

Last Answer : Multiple of 3 on a die = 3, 6 ∴ P (a multiple of 3) = 2/6 = 1/3.

Description : Two dice are rolled simultaneously. The probability of getting a multiple of 2 on one dice and a multiple of 3 on the other is -Maths 9th

Last Answer : (c) \(rac{11}{36}\)Total number of outcomes when two identical dice are rolled, n(S) = 6 6 = 36 Let A : Event of rolling a multiple of 2 on one die and a multiple of 3 on the other die ⇒ A = {(2, 3), (2, 6), (4, 3), (4, ... , 4), (3, 6)} ⇒ n(A) = 11 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{11}{36}\).

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).

Description : Two dice are rolled once. Find the probability of getting an even number on the first die, or a total of 7. -Maths 9th

Last Answer : (c) \(rac{7}{12}\)Total number of ways in which 2 dice are rolled = 6 6 = 36 ⇒ n(S) = 36 Let A : Event of rolling an even number of 1st dice B : Event of rolling a total of 7 ⇒ A = {(2, 1), (2, 2) , (2, 6), (4 ... (rac{18}{36}\) + \(rac{6}{36}\) - \(rac{3}{36}\) = \(rac{21}{36}\) = \(rac{7}{12}\).

Description : When two dice are thrown, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 9. A) 1/2 B) 1/5 C) 2/5 D) 4/2 

Last Answer : Answer: A) Let the event of getting a greater number on the first die be G. There are 4 ways to get a sum of 9 when two dice are rolled = {(3,6),(4,5),(5,4), (6,3)}. And there are two ways where the number on the ... Now, P(G) = P(G sum equals 9)/P(sum equals 9) = (2/36)/(4/36) = 2/4 =>1/2

Description : A dice is thrown once, what is the probability of getting a prime number? a. 1/3 b. 6/25 c. 1/2 d. 1/4

Last Answer : c. 1/2

Description : In a single throw of two dice, what is the probability of getting a sum of 9? -Maths 9th

Last Answer : Outcomes with sum of 9 = { (3, 6), (4, 5), (5, 4), (6, 3) } P ( getting a sum of 9 is ) = 4/36 = 1/9

Description : A die is thrown twice. What is the probability that at least one of the two throws comes up with the number 3 ? -Maths 9th

Last Answer : (b) \(rac{11}{36}\)Let S = total ways in which two dice can be rolled ⇒ n(S) = 6 6 = 36 Let A : Event of throwing 3 with 1st dice, B : Event of throwing 3 with 2nd dice. Then, A = {(3, 1), (3, 2), (3, 3), (3, 4), ... ) - P(A ∩ B)= \(rac{6}{36}\) + \(rac{6}{36}\) - \(rac{1}{36}\) = \(rac{11}{36}\).

Description : Two dice are thrown simultaneously 500 times. -Maths 9th

Last Answer : (i) P (getting a sum more than 10) = P (getting a sum of 11) + P (getting a sum of 12) = 28/500 + 15/500 = 28 + 15/500 = 43/500 = 0.869 = 0.09 (ii) P (getting a sum less than or equal to 5) = P ( ... + P (getting a sum of 10) + P (getting a sum of 11) = 53/500 + 46/500 + 28/500 = 127/500 = 0.254

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

Description : Two dice are thrown together. The probability that the total score is a composite number is: A) 5/12 b) 12/7 c) 7/12 d) 12/5

Last Answer :  Answer: C)  Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a composite number Then E= { (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 5), (3, 3), (3, 6), (4, 2), (4,4),(4, 5), (4, 6), ( ... 5,3),(5,4),(5,5),(6,2),(6,3),(6,4),(6,6) } n(E) = 21 P(E) = n(E)/n(S) = 21/36 = 7/12.

Description : Two dice are thrown together. Find the probability that the sum of the numbers obtained is even a. 1/4 b. 1/6 c. 1/3 d. 1/2

Last Answer : d. 1/2

Description : When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8. a) 5/18 b) 1/18 c) 2/5 d) 1/5

Last Answer : c) 2/5

Description : In a throw of a die, find the probability of getting an even number. -Maths 9th

Last Answer : Total even number on a die = 3 P (getting an even numbers) = 3/6 = 1/2

Description : A die is rolled three times. The probability of getting a larger number than the previous number each time is: -Maths 9th

Last Answer : (b) \(rac{5}{24}\)Total number of ways three die can be rolled = 6 6 6 = 216 A larger number than the previous number can be got in the three throws as (1, 2, 3), (1, 2, 4), (1, 2, 5) ( ... , 5, 6). ∴ Total number of favourable cases = 20∴ Required probability =\(rac{20}{216}\) = \(rac{5}{24}\).

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

Description : In a throw of a die, find the probability of not getting 4 or 5. -Maths 9th

Last Answer : Required probability = 1 – P(4) – P(5) =1- 1 / 6 - 1 / 6 = 4 / 6 = 2 / 3

Description : What is the probability of getting a 4 or a 6 when a die is thrown together? a) 2/3 b) 1/3 c) 3/6 d) 4/6

Last Answer : Answer: B) Taking the individual probabilities of each number, getting a 4 is 1/6 and so is getting a 6. Applying the formula of compound probability, Probability of getting a 4 or a 6, P(4 or 6) = P(4) + P(6) – P(4 and 6) ==> 1/6 + 1/6 – 0 2/6 = 1/3

Description : A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd is -Maths 9th

Last Answer : (a) \(rac{1}{2}\) Let S be the sample spaceThen n(S) = Number of four digit numbers that can be formed with out repetition1 of 4 digits1 of 3 digits1 of 2 digits1 of 1 digitThHTO= 4! = 4 3 2 1 ways = 24 ways. Let E : ... 1 = 12∴ P(E) = \(rac{n(E)}{n(S)}\) = \(rac{12}{24}\) = \(rac{1}{2}\).

Description : A die is thrown six times and number on it is noted as given below : -Maths 9th

Last Answer : Here, in 6 trials, each number occur once and total prime numbers i.e., 2, 3, 5 occur one time each Hence, the number of prime occur = 3 Probability of getting a prime = 3/6 =1/2

Description : A die is thrown six times and number on it is noted as given below : -Maths 9th

Last Answer : Here, in 6 trials, each number occur once and total prime numbers i.e., 2, 3, 5 occur one time each Hence, the number of prime occur = 3 Probability of getting a prime = 3/6 =1/2

Description : Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die ? (A) 22 / 36 (B) 12 / 36 (C) 14 / 36 (D) 6 / 36

Last Answer : (C) 14 / 36 

Description : A coin and six faced die, both unbiased are thrown simultaneously. -Maths 9th

Last Answer : (c) \(rac{1}{4}\)Let A : Event of getting a tail on the coin B : Event of getting an even number on the die. Then, P(A) = \(rac{1}{2}\)P(B) = \(rac{3}{6}\) = \(rac{1}{2}\) as B = {2,4,6}A and B being independent events ... die)= P(A ∩ B) = P(A) P(B) = \(rac{1}{2}\)x\(rac{1}{2}\) = \(rac{1}{4}\).

Description : When a dice is rooled, find the probability of getting an even prime number.

Last Answer : When a dice is rooled, find the probability of getting an even prime number. A. `1/6` B. `1/3` C. `1/2` D. `5/6`

Description : If a dice is rooled, then the probability of getting a prime number is _______.

Last Answer : If a dice is rooled, then the probability of getting a prime number is _______.

Description : A dice is rolled 250 times, and the outcoms 1, 2, 3, 4, 5 and 6 occurred as given in the following table: Find the probalility of getting an odd numbe

Last Answer : A dice is rolled 250 times, and the outcoms 1, 2, 3, 4, 5 and 6 occurred as given ... table: Find the probalility of getting an odd number.

Description : A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Last Answer : (b) \(rac{11}{15}\)n(S) = 300 Let A : Event of getting a number divisible by 2 B : Event of getting a number divisible by 3 C : Event of getting a number divisible by 5 ∴ A ∩ B : Event of getting a number divisible by ... \(rac{320}{300}\) - \(rac{100}{300}\) = \(rac{220}{300}\) = \(rac{11}{15}\).

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

Description : A dice is rolled number of times and its outcomes are recorded as below : -Maths 9th

Last Answer : Total number of outcomes = 250 Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138 P (getting an odd number) = 138 / 250 = 69 / 125

Description : When a dice is thrown, the outcomes are ______.

Last Answer : When a dice is thrown, the outcomes are ______.

Description : Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails. -Maths 9th

Last Answer : (d) \(rac{1}{2}\)Let S be the sample space. Then, S = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT} ⇒ n(S) = 8 Let A : Event of getting more heads than number of tails. Then, A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{8}\) = \(rac{1}{2}.\)

Description : If a dics is thrown, then the probability of getting an even number is _______.

Last Answer : If a dics is thrown, then the probability of getting an even number is _______.

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Description : If two coins are tossed once, what is the probability of getting at least one head ? -Maths 9th

Last Answer : When two coins are tossed once, there are four possible outcomes, i.e., S = {HH, HT, TH, TT} ∴ Total number of outcomes = n(S) = 4 Let A : Event of getting at least one head ⇒ A = {HH, HT, TH} ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{4}.\)

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Description : What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th

Last Answer : A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, ... ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{7}.\)

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

Description : Two dice are rolling simultaneously .What is the probability that the sum of the number on the two faces is divided by 5 Or 7. A) 13/36 B) 14/36 C) 11/36 D) 9/36 

Last Answer :  Answer: C) Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 5or 7. Then,E = {(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(4,6),(5,2),(5,5),(6,1),(6,4)} n(E) = 11. Hence, P(E) = n(E)/n(S) = 11/36

Description : A dice is rolled, the probability that the number on the face showing up is greater than 6 is ______.

Last Answer : A dice is rolled, the probability that the number on the face showing up is greater than 6 is ______.

Description : Now it was Ravi‘s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8? a. 1 b. 5/36 c. 1/18 d. 5/18

Last Answer : d. 5/18

Description : Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7? a. 5/9 b. 5/36 c. 1/6 d. 0

Last Answer : c. 1/6

Description : Now it was Ravi‘s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12? a. 1 b. 5/36 c. 1/18 d. 0

Last Answer : a. 1

Description : Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13? a. 1 b. 5/36 c. 1/18 d. 0

Last Answer : d. 0

Description : What is the probability if a fair dice rolled on 2?

Last Answer : Assuming a standard 6-sided die with sides numbered {1, 2, 3, 4,5, 6}, then:Number of successes = 1Total number of outcomes = 6→ pr(2) = 1/6

Description : If you were spin a spinner numbered 1-10 what would is the probability of getting an odd number?

Last Answer : It is 0.5

Description : Find the probability that a two digit number formed by the digit 1, 2, 3, 4 and 5 is divisible by 4. -Maths 9th

Last Answer : The two digit numbers can be formed by putting any of 5 digits at the one 's place and also one of the 5 digits at ten's place. So, Total number of 2-digit numbers that can be formed using these 5-digits = 5 5 = ... 52}, i.e, 5 in number. ∴ Required probability = \(rac{5}{25}\) = \(rac{1}{5}.\)

Description : Can the experimental probability of an event be a negative number? -Maths 9th

Last Answer : No, because the number of trials in which the event can happen cannot be negative and the total number of trials is always positive.