Login
Get Advice on Live Video Call
Earn $ Cash $ with
consultations on Bissoy App

Numbers are selected at randm, one at time, from the two-digit numbers `00,01,02…,99` with replacement. An event E occurs if the only product of the t

Monjur Alahi

Numbers are selected at randm, one at time, from the two-digit numbers `00,01,02…,99` with replacement. An event E occurs if the only product of the two digits of a selected number is 18. If four members are selected, find the probability that the event E occurs at least 3 times.


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - `(97)/((25)^(4))`
The given numbers are `00,01,02…,99.` There are total 100 numbers, out of which the numbers, the product of whose digits is 18, are `29,36,63,and 92.`
`thereforep=P(E)=(4)/(100)=1/25`
` impliesq=1-p =24/25`
From binomial distribution,
P(E occuring at least 3 times)
=P(E occuring 3 times)+ P (E occuring 4 times)
`=""^(4)C_(3)p^(3)q+""^(4)C_(4)p^(4)`
`=4xx((1)/(25))^(3)((24)/(25))+((1)/(25))^(4)=(97)/((25)^(4))`

0 like 0 dislike
1 views
A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, then what is the probab

Probability of defective watch from a lot of 100 watches=`10/100=1/10` `thereforep=1//10,q=9/10,n=8and rge1` `thereforeP(rge1)=1-P(r=0)=1-""^(8)C_(0)(1/10)^(0)(9/10)^(8-0)` `=1-(8!)/(0!8!)cdot(9/10)^(8)=1-(9/10)^(8)`

2 Answers 1 views
A flashlight has 8 batteries out of which 3 are dead. If two batteries selected without replacement and tested, then probability that both are dead is

Required probability=`P_(D)cdotP_(D)=3/8cdot2/7=3/28`

2 Answers 1 views
`Aa n dB` are two candidates seeking admission in ITT. The probability that `A` is selected is 0.5 and the probability that `Aa n dB` are selected is

Correct Answer - Not possible P(A) = 0.5, `P(A nn B) le 0.3` So, `P(A uu B) = P(A) + P(B) - P(A nn B)` gives `P(B) = P(A uu B)...

2 Answers 1 views
A three-digit number is selected at random from the set of all three-digit numbers. The probability that the number selected has all the three digits

Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...

2 Answers 1 views
The probability of happening an event A in one trial is 0.4. Find the probability that the event A happens at least one in three independent trials.

Correct Answer - `4.0.784` Here `P(A)=0.4and P(barA)=0.6` Probability that A does not happen at all is `(0.6)^(3).` Thus, required probability is `1-(0.6)^(3)=0.784.`

2 Answers 1 views
One ticket is selected at random from 100 tickets numbered `00,01,02, …, 99.` Suppose A and B are the sum and product of the digit found on the ticket

Correct Answer - B The sum of the digits can be 7 in the following ways: `07,16,25,34,43,52,61,70.` `therefore(A=7)={07,16,25,34,43,52,61,70}` Similarly, `(B=0)={00,01,02...,10,20,30,...,90}` Thus, `(A=7)nn(B=0)={09,70}` `thereforeP((A=7)nn(B=0))=2/100,P((B=0))=19/100` `P(A=7)|B=0=(P((A=7)nn(B=0)))/(P(B=0))` `=(2/100)/(19/100)=2/19`

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. Which one of the following event

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.

2 Answers 1 views
One ticket is selected at ransom form 50 tickets numbered `00,01,02,…,49.` Then the probability that the sum of the digits on the selected ticket is 8

Correct Answer - A `S={00,01,02,....,49}` Let A be the event that the sum of the digits on the selected ticket is 8. Then `A={08,17,26,35,44}` Let B be the event that the...

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
Two numbers 'a' and 'b' are selected successively without replacement in that order from integers 1 to 10.

We have a set of natural numbers from 1 to 10 where a and b are two variables which can take values from 1 to 10. So, total number of possible...

2 Answers 1 views

Log in to ask questions, provide answers, or leave comments.


Login

Install the Bissoy app to consult with a doctor.


Download