Pls Help !!The chance that a seed will germinate is 7/9. There are 9 seeds sown.

Question

Given that at least 6 seeds will germinate, what is the probability that

(i) fewer than 8 seeds will germinate?

(ii) more than 7 seeds will germinate?

Answer 1

Hi Cheukyiu:

Let's solve together.

We use the Binomial Distrib: Prob(r successes, with prob = p, from n trials) = [n!/r! (n-r)!] p^r q^n-r

Here, we have p = 7/9, q = 2/9 (the probability of NOT germinating =1 - 7/9), and n = 9.

In question (i), we want the probability that 7 seeds germinate, that is, r = 7 .

So P(7 seeds germinate) = (9!/(7! 2!) (7/9)⁷ x (2/9)²

= (9 x 8) / 2 x (7/9)⁷ x (2/9)²

= 0.0.306 = 30.6%

In question (ii), we want the prob that 8 or 9 seeds germinate. These are mutually exclusive events (either 8 germinate or 9, but not both) so we can add the individual probabilities together: P(8 or 9) = P(8) + P(9)

P(8 germinate) = 9! / (8! 1!) x (7/9)⁸ x (2/9)¹ = 2 x (7/9)⁸ = 0.268 = 26.8%

P(9 germinate) = 9! / (9! 0!) x (7/9)⁹ x (2/9)⁰ = 1 x (7/)⁹ x 1 [Remember that 0! = 1 and (2/9)⁰ = 1 ]

= 0.104 = 10.4%

Therefore, P(8 or 9 germinate) 26.8% + 10.4% = 37.2%

I hope this helps!