According to the manufacturer, about 26% of sour candy in a package of sandy's sours are grape. what is the probability that the first grape candy chosen from the bag will be, at least, the third candy chosen overall?

Step-by-step explanation:

We assume that there are 100 sour candies, Thus-  

26 % of candy are grape implies that 26% of 100 candies are grape that is equal to 26

Now remaing candies that are not grape are 100-26 = 74

Based on the rule of multiplication:

P(A ∩ B) = P(A)/ P(B|A)  

In the beginning, there are 26grape candies, probability of choosing first grape candy =  26C1 = 26

After the first selection, we replace the selected grape candy so there are still 100 candies in the bag P(B|A) =  100C3 = 100 x 99 x 98 x 97!/3! X 97!

= 50 x 33 x 98

So probability =1/ 50 x 33 x 98 x 26

= 1/4204200

1/ 50 x 33 x 98 x 26

= 1/4204200

Step-by-step explanation:1/ 50 x 33 x 98 x 26

= 1/4204200

step-by-step explanation:

i am not sure sorry : c

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47