Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i. Which one of the following event

Monjur Alahi

Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i.
Which one of the following events is most probable?
A. `A_(3)`
B. `A_(4)`
C. `A_(5)`
D. `A_(6)`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A
`P(A_(2))=18/36,`
`P(A_(3))=12/36=1/3,`
`P(A_(4))=9/36=1/4`
`P(A_(5))=7/36`
`P(A_(6))=6/36=1/6`
Hence, `A_(3)` is most probable.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Two identical fair dice have numbers 1 to 6 written on their faces. Both are tossed simultaneously. What is the probability that the product

Correct option is: B. \(\frac{1}{9}\)

2 Answers 44 views
There are two die A and B both having six faces. Die A has three faces marked with 1, two faces marked with 2, and one face marked with 3. Die B has o

Correct Answer - C x can be 2, 3, 4, 5, 6. The number of ways in which sum of 2, 3, 4, 5, 6 can occur is given by the...

2 Answers 2 views
There are two die A and B both having six faces. Die A has three faces marked with 1, two faces marked with 2, and one face marked with 1, two faces m

Correct Answer - C The number of ways in which different sums can occurs is (3 + 2 + 1)(1 + 2 + 3) = 36. The probability of 4 is...

2 Answers 1 views
A dice is weighted such that the probability of rolling the face numbered n is proportional to `n^(2)` (n = 1, 2, 3, 4, 5, 6). The dice is rolled twic

`P(n) = Kn^(2)` Given `P(1) = K, P(2) = 2^(2)K, P(3) = 3^(2)K, P(4)= 4^(2)K`, `P(5) = 5^(2)K, P(6) = 6^(2)K` So, total probability = 91K implies 1 = 91K...

2 Answers 1 views
Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i. For which one of the following (

Correct Answer - C `P(A_(2))=1/2,P(A_(3))=1/3,P(A_(6))=1/6` `thereforeP(A_(2)nnA_(3))=P(A_(2))P(A_(3))` `impliesP(A_(6))=P(A_(2))P(A_(3))` `6/36=1/2xx1/3` Hence, `A_(2)and A_(3)` are independent.

2 Answers 1 views
Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i. The number of all possible order

Correct Answer - D Note that `A_(1)` is independent with all events `A_(1),A_(2),A_(3),A_(4)....,A_(12),` Now, total number of ordered pairs is 23. `underset(22)( underbrace((1.1),(1,2),(1.3),...,(1,11)))+(1,12)` Also that `A_(2),A_(3),andA_(3),A_(2)` are independent. Hence, there are...

2 Answers 3 views
If two loaded dice each have the property yjsy 2 or 4 is there times as likely to appears as `1,3,5,or6` on each roll. When two such dice are rolled,

Correct Answer - 7 Let the probaility of the faces `1,3,5` or 6 be p for each face. Hence, probability of each of the faces 2 or 4 is 3p Now...

2 Answers 1 views
A fair coin is flipped n times. Let E be the event "a head is obtained on the first flip" and let `F_(k)` be the event "exactly k heads are obtained".

Correct Answer - 2 Here `P(E)=1/2and P(F_(k))=""^(n)C_(k).(1)/(2^(n))` Also, `P(EnnF_(k))=p` (exactly k heads are obtained and head obtained in first filp) `=1/2""^(n-1)C_(k-1)((1)/(2))^(n-1)` Events E and `f_(k)` are independent. Therefore, `P(EnnF_(k))=P(E).P(F_(k))` `or ""^(n-1)C_(k-1)xx(1)/(2^(n))=1/2xx""^(n)C_(k)(1)/(2^(n))`...

2 Answers 3 views
Let two fari six-faced dice A and B be thrown simltaneously. If `E_(1)` is the event that die A shows up four, `E_(2)` is the event that die B shows u

Correct Answer - C `E_(1):{(4,1),...,(4,6)}to6"cases"` `E_(2):{(1,2),...,(6,2)}to6"cases"` `E_(3):"18 cases (sum of both is odd")` `thereforeP(E_(1))=3/6=1/6=P(E_(2))` `P(E_(3))=18/36=1/2` `P(E_(1)nnE_(2))=1/36` `P(E_(2)nnE_(3))=3/36=1/12` Similarly `P(E_(3)nnE_(1))=1/12` `P(E_(1)nnE_(2)nnE_(3))=0` `therefore E_(1),E_(2),E_(3)` are not independent.

2 Answers 1 views
The probability of event `A` is `3//4`. The probability of event `B`, given that event `A` occurs is `1//4`. The probability of event `A`, given that

Correct Answer - C `(c )` `P(A)=(3)/(4)`, `P(B//A)=(1)/(4)` `P(A//B)=(2)/(3)` `P(B//A)=(P(BnnA))/(P(A))=(1)/(4)` (given) `:.P(BnnA)=(3)/(4)*(1)/(4)=(3)/(16)` Now `P(A//B)=(P(AnnB))/(P(B))=(2)/(3)`(given) `:.P(B)=(3)/(2)*(3)/(16)=(9)/(32)` `:. P(AuuB)=(3)/(4)+(9)/(32)-(3)/(16)=(24+9-6)/(32)=(27)/(32)` `:. P(A^(C )nnB^(C ))=1-P(AuuB)=1-(27)/(32)=(5)/(32)`

2 Answers 1 views
X