find the probability of throwing at most 2 sixes in 6 throws of a single die. - random variables and its probability distributions
find the probability of throwing at most 2 sixes in 6 throws of a single die.
let x denote the number of sixes. p(getting a six when a die is thrown) = p = `(1)/(6)` ∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)` given, n = 6∴ x ∼ b`(6, 1/6)`the p.m.f. of x is given by p(x = x) = `""^6"c"\_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6p(getting at most 2 sixes) = p(x ≤ 2) = p(x = 0 or x = 1 or x = 2) = p(x = 0) + p(x = 1) + p(x = 2) = `""^6"c"\_0(1/6)^0 (5/6)^6 + ""^6"c"\_1(1/6)(5/6)5 + ""^6"c"\_2(1/6)^2 (5/4)^4` = `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4` = `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]` = `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]` = `(5/6)^4 (70/36)` = `(5/6)^4 (35/18)` = `(5/4)^4 ((5 xx 7)/(6 xx 3))` = `(7)/(3)(5/6)^5`.
Total no. of possible outcomes = 30 {1, 2, 3, … 30}  

  

E ⟶ event of getting no. divisible by 3.  

  

No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}  

  

Probability, P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)  

  

P(E) = 10/30 = 1/3  

  

Bar E ⟶ event of getting no. not divisible by 3.  

  

P(Bar E) = 1 – P(E)  

  

= 1 - 1/3 = 2/3 given : five defective bulbs are accidently mixed with 20 good ones.to find : probability distribution from this lotformula used :five defective bulbs are accidently mixed with 20 good ones.total number of bulbs = 25x denote the number of defective bulbs drawnx can draw the value 0 , 1 , 2 , 3 , 4.since the number of bulbs drawn is 4, n = 4p(0) = p(getting a no defective bulb) = = = p(1) = p(getting 1 defective bulb and 3 good ones) = = p(1) = = p(2) = p(getting 2 defective bulbs and 2 good one) = p(2) = = = p(3) = p(getting 3 defective bulbs and 1 good one) = p(3) = = p(4) = p(getting all defective bulbs) = = = p(4) = the probability distribution table is as follows, and what about guessing? are you good at guessing? try to guess if i am or not, if you guess right you choose your reward ;p ?answer :




step 1 of 6:


given, in a sample of 20 weld failures, 85% of them occur in the weld metal itself, and the remaining 15% occur in the base metal.


let x follows binomial distribution with probability mass function is




p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


where,
n = sample size
 = 20
x = random variable
p = probability of success
 = 15% 
 = 0.15
q = 1 - p (probability of failure)


mean of the binomial distribution is 


 = np


variance of the binomial distribution is 


 = npq


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step 2 of 6:




1. the claim is to find the probability that exactly five of them are base metal failures


we have to find p(x=5) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 5


p(x = 5) = (1 - 0.15


 = (15504) (0.0000759) (0.087354)


 = 0.1028


hence, the probability that exactly five of them are base metal failures is 0.1028.


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step 3 of 6:




1. the claim is to find the probability that fewer than four of them are base metal failures


we have to find p(x < 4) = p(x = 0) + p( x = 1) + p(x = 2) + p(x = 3) 






p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0, 1, 2 and 3


p(x < 4) = (1 - 0.15+ (1 - 0.15+


 (1 - 0.15+ (1 - 0.15


 = (1) (1) (0.03876) + (20) (0.15) (0.0456) +
 (190) (0.0225) (0.053646) + (1140) ( 0.003375) (0.0631134)


 = 0.03876 + 0.1368 + 0.2293 + 0.2428


 = 0.6477


hence, the probability that fewer than four of them are base metal failures 0.6477


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step 4 of 6:




c) the claim is to find the probability that none of them are base metal failures.


we have to find p(x=0) 


p(x: n,p) = (1 - p , x = 0, 1, 2, ….., n.


 where, n = 20, p = 0.15 and x = 0


p(x = 0) = (1 - 0.15


 = (1) (1) (0.03876) 


 = 0.03876


hence, the probability that none of them are base metal failures is 0.03876


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step 5 of 6:




d) the claim is to find the mean number of base metal failures..


 mean of the binomial distribution is = np


where, n = 20, and p = 0.15


substitute these values in 


then, = (20) (0.15)


 = 3


hence, the mean number of of base metal failures is 3.


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step 6 of 6:




e) the claim is to find the standard deviation of the number of base metal failures..


standard deviation of the binomial distribution is 


where, n = 20, p = 0.15 and q = 1 - p


 = 1 - 0.15


 = 0.85


substitute these values in 


then, 


 


 = 1.596


hence, the standard deviation of the number of base metal failures is 1.596.


\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is . several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 start with 50 easys. if you solve those quickly, then move into mediums. hards are pretty tricky tbh i just attempt then peek answers/memorize lol a) Let's find z - score:  
z = (x-mu)/sigma = (40,000-47,000)/4,000 = - 1.75  
p-value for z = 1.75 is 0.0401, for estimating the percent of the tires which last for at least 40,000 miles we have to consider right side of the distribution:  
P(x > 40,000) = (1-P(x < 40,000))*100% = 95.99 %  
  
  
b) the percent of the tires which are not last for at least 40,000 miles would be equal to P (x < 40,000) = 0.0401*100% = 4.01 %;  
c) for more than 50,000 miles   
z = (50,000-40,000)/4,000 = 2.5  
the probability would be  
P(x>50,000) = 00.62 % have sex tonight.<br><br>best wishes total number of cards in a well-shuffled pack =52total number of queens and kings cards =4+4=8therefore, there are 44 cards (52−8) that are neither king nor queen.total number of favorable outcomes  =44  
probability is given by,p(a)=no. of total outcomesno. of possible outcomes​  
required probability =5244​=1311​