# Binomial DistributionIt is known that a particular variety of lettuce seed has a probability of 0.85 of successfully producing a lettuce. Find the probability that if 200 such seeds are sewn, the...

Binomial Distribution

It is known that a particular variety of lettuce seed has a probability of 0.85 of successfully producing a lettuce. Find the probability that if 200 such seeds are sewn, the number of seeds produced will be less than 160?

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Take note of the formula:

P(x = k) = nCk*p^k*(1-p)^(n-k)

where n = number of trials

x = number of successes

p = probability of success of a n individual trial

Here, take note of the word "less than", so, we will consider x = 0,1,2,3,4,5,6,...159. But that will be tedious, we can use another formula: P(x < k) = 1 - P(x greater than or equal to k). This is called Complement Rule.

So, we will solve consider x = 160,170,171,172,173,...200. And we will get the sum of the probability of those, and subtract it from 1.

Considering P(x = 160) = 200C160 * (0.85)^160 * (0.15)^40= 0.0115469107

P(x=161) = 200C161 * (0.85)^161 * (0.15)^39= 0.0162565203

P(x=162) = 200C162 * (0.85)^162 * (0.15)^38= 0.0221771048

P(x=163) = 200C163* (0.85)^163 * (0.15)^37= 0.029973614

P(x=164) = 200C164* (0.85)^164 * (0.15)^36= 0.0374553664

P(x=165) = 200C165 * (0.85)^165 * (0.15)^35= 0.0463084531

P(x=166) = 200C166* (0.85)^166 * (0.15)^34= 0.0553283726

P(x=167) = 200C167 * (0.85)^167* (0.15)^33= 0.0638319349

P(x=168) = 200C168 * (0.85)^168* (0.15)^32= 0.0710510228

P(x=169) = 200C169 * (0.85)^169 * (0.15)^31= 0.0762362059

Continuing the process up to P(x = 200),

Adding the results we will have:

P(x greater than or equal to 160) = .9664538146 or 0.9665.

So, P( x < 160) = 1 - 0.9665 = 0.0335

We can also use our calculator, (like TI-84 Plus),

Press 2nd --> VARS(DISTR) --> B:binomcdf. Enter trials: 200

p: 0.85

x value: 160

then move the cursor to Past, and press ENTER twice.

-Glad to help:)