Genetic Probability Kahn Academy Questions HELP

[Radioactive1112]

Suppose the brown allele for eye color (B) is completely dominant over the blue allele for eye color (b). If two brown-eyed parents produce a child that is blue-eyed, what is the probability that at least one out of the next two children they produce will also have blue eyes?

• 44%
• 50%
• 25%
• 12.5%

Im confused on what the answer should be...I drew a punnett square and bb (blue eyed) will be produced 25% of the time. So at least one out of the next two children would have blue eyes right? Kahn says the answer is 44%....I dont know if this is a glitch or am I am going about this the wrong way....help please.

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[NeuroMaster316]

Be careful, at least one out of the next two means either one or BOTH will have blue eyes i.e. the 25% refers to the probability of one child having blue eyes. In these sort of Qs, to save time, just calculate the probability of the opposite not happening. Since 1/4 will have blue eyes, then 3/4 will have brown ones.
Both children having brown eyes (zero blue) is 3/4 x 3/4
So the opposite would be 1 - (3/4 x 3/4) which equals to about 44%
To simplify your life, when you see AT LEAST always calculate the probability of obtaining the complement then deduct it from 1

EDIT: To clarify about the other (harder way), you'd add the probability of only the 1st kid being blue, only the 2nd being blue, and both being blue:
0.25x0.75 + 0.75x0.25 + 0.25x0.25 = 0.4375 = 0.44

 

[efle]

This is a classic 1 - 6 - 9, it's always the same for two heterozygous mating

Both Blue 1/16 (1/4 * 1/4)
One Blue 6/16 (3/4 * 1/4 + ditto in opposite order)
No Blue 9/16 (3/4 * 3/4)

Odds of at least 1 = 1/16 + 6/16 = 7/16 = 44%

 

[Labrat07]

I don't understand these calculations. From my statistic understanding.

For OR you add the ratio. In this case, if only one of two kids have blue eye, I thought it'd be
1/4 +1/4 = 2/4.

For AND you multiply. In this case,
1/4 * 1/4 = 1/16.

I need some clarifying. Please someone teach me another way to understand

 

[NeuroMaster316]

I don't understand these calculations. From my statistic understanding.

For OR you add the ratio. In this case, if only one of two kids have blue eye, I thought it'd be
1/4 +1/4 = 2/4.

For AND you multiply. In this case,
1/4 * 1/4 = 1/16.

I need some clarifying. Please someone teach me another way to understand

You want at least one of the kids to have blue eyes, so you have three options: either the 1st one is blue eyed, the 2nd is blue eyed, or both are blue eyed so you add the three probabilities.

If only one is blue eyed, then the other is brown eyed so what is the probability of having a blue eyed kid and the other one being brown eyed ? It's 1/4 x 3/4 = 3/16
What is the probability of having one kid blue eyed and the other one blue eyed? It's 1/4 x 1/4 = 1/16
Now add the probabilities : 3/16 + 3/16 + 1/16 = 7/16

Alternatively you can deduct the complement from 1.
The complement of AT LEAST events is NONE i.e. what is the probability of neither kid being blue eyed = both being brown eyed? It's 3/4 x 3/4 = 9/16
The probability of at least one being blue eyed is therefore 1 - 9/16 = 7/16

 

[Labrat07]

@NeuroMaster316 I understand now. Thank you so much.

 

[Radioactive1112]

Be careful, at least one out of the next two means either one or BOTH will have blue eyes i.e. the 25% refers to the probability of one child having blue eyes. In these sort of Qs, to save time, just calculate the probability of the opposite not happening. Since 1/4 will have blue eyes, then 3/4 will have brown ones.
Both children having brown eyes (zero blue) is 3/4 x 3/4
So the opposite would be 1 - (3/4 x 3/4) which equals to about 44%
To simplify your life, when you see AT LEAST always calculate the probability of obtaining the complement then deduct it from 1

EDIT: To clarify about the other (harder way), you'd add the probability of only the 1st kid being blue, only the 2nd being blue, and both being blue:
0.25x0.75 + 0.75x0.25 + 0.25x0.25 = 0.4375 = 0.44

ooooooo.....I see now. Yes, the method you mentioned is much easier. Thank you

 

[krukshanks]

Classic conditional probability question. Check this out if you want to know more about the systematic information behind the calculations:
http://en.wikipedia.org/wiki/Conditional_probability