# Probability of certain ratio

#### WMDhamnekar

According to the theory of genetics a certain cross of guinea pigs will result in red, black and white offspring in the ratio 8:4:4. Find the probability that among 8 such offspring 5 will be red, 2 black and 1 white.

Answer provided is 0.082. I computed the answer in this way $\frac58*\frac24*\frac14=0.078125$ How did the author compute 0.082?

If any member want to clarify my query by replying to this question, I welcome all member's correct replies.

#### Debsta

MHF Helper
I think you're trying to say that the probability of getting 5 reds out of 8 is 5/8. That is incorrect.

R:W:B = 8:4:4 = 2:1:1

Therefore P(R) =2/4 = 0.5, P(W)=1/4 =0.25, P(B)=1/4 =0.25.

Can you continue?

Think of the number of ways you can get the guinea pigs eg RRRRRWWB, RRRRWWRBB, etc. This will be used in your calculation of the probability required.

#### WMDhamnekar

I think you're trying to say that the probability of getting 5 reds out of 8 is 5/8. That is incorrect.

R:W:B = 8:4:4 = 2:1:1

Therefore P(R) =2/4 = 0.5, P(W)=1/4 =0.25, P(B)=1/4 =0.25.

Can you continue?

Think of the number of ways you can get the guinea pigs eg RRRRRWWB, RRRRWWRBB, etc. This will be used in your calculation of the probability required.
Hello,
But how did the author compute 0.082. ? My computation is from the 8 offsprings, the probability that any 4 would be red is 0.5.So $\binom{8}{4}$=70 is 50% of the total ways to select the ratio 8:4:4. So how to proceed further?

#### romsek

MHF Helper
I don't understand this problem. It states the proportions of offspring color are 8:4:4 and then asks for the probability of a triplet that doesn't adhere to those proportions.
What does it mean for the proportions to be what they are? Are those the mean proportions, i.e. that in the long run

$p(r)=\dfrac 1 2,~ p(b)=\dfrac 1 4,~ p(w)=\dfrac 1 4$ ?

Oh, I see this is exactly what Debsta has proposed. Continue with his advice.
How many ways can you choose 5 from 8? Then there are 3 left that you must choose 2 from.
If the probability of 1 red is $p_r$ what is the probability of getting red 5 times (assuming independence) ?

If that's the case the probability of $(5,2,1)$ should be

$p =\dbinom{8}{5}\dbinom{3}{2}\left(\dfrac 1 2\right)^5 \left(\dfrac 1 4\right)^2 \left(\dfrac 1 4\right)^1 = \dfrac{21}{256} \approx 0.082$

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#### Debsta

MHF Helper
Oh, I see this is exactly what Debsta has proposed. Continue with his advice.