3) The chance that a salmon jumps up the waterfall to upstream successfully 35%. We have a group of 9 salmon. What is the chance that none of them jump up successfully? What is the chance that all of them reach up-stream? What is the chance that at least 2 of them reach up-stream? What is the chance that at most 1 of them reaches up-stream? Is “at least 1” complement of the event “at most 1”?

(a) P(0,9) = 0.0207

(b) P(2-9,9) = 0.1004

(c) P(0-1,9) = 0.1211

(d) "at least 1" and "at most 1" are not complements of each other because there is an overlap of "1" in both cases.

Step-by-step explanation:

With appropriate assumptions, this can be solved using the binomial distribution.

Probability of success for each trial, p = 35%

Number of trials, n = 9

using formula for x successes out of n trials, each with probability p

P(x,n) = C(n,x) p^x (1-p)^(n-x)

where C(n,x) = n!/(x!(n-x)!)

(a) zero success

n=9

p=0.35

x = 0

P(0,9) = 9!/(0!9!) 0.35^0 * 0.65^9

= 1*1* 0.0207

= 0.0207

(b) 2 or more successes

We need the sum of probabilities of 2,3,4,5,6,7,8,9 successes, which is easier calculated by (1-P(0,9)-P(1,9))

P(1,9) = C(9,1) * p^1 * p^8

= 9 * 0.35 * 0.65^8

= 9 * 0.35 * 0.3186

= 0.1004

Therefore

P(2 to 8, 9)

= 1 - P(0,9) - P(1,9)

= 1 - 0.0207 - 0.1004

= 0.8789

(c) at most 1 success

P(0,9) + P(1,9)

= 0.0207 + 0.1004

= 0.1211

take the -3 and add it to the 9 so it will make 12, so then ur left with 4x=12 divide the 4 from x by 12 and then you will be left with x=3