(i) Choosing the first girl with blue eyes = 3/10. Since there is no replacement, the probability of the second girl having blue eyes = 2/9. The probability of both/all occurring is found by multiplying the individual probabilities.
(ii) Do similar process with non-blue probabilities.
(iii) This is either one or two have blue eyes. You have probability of two from (i). Probability of one = 3/10 * 7/9 or 7/10 * 3/9. (note that they have the same product). Either-or probability is found by adding the individual probabilities.
Of 10 girls in a class, 3 have blue eyes. If two girls are chosen at random, what is the probability that (i) both have blue eyes, (ii) neither has blue eyes, (iii) at least one has blue eyes?
I have the answers but do not know how to get them. I'm trying to use a tree diagram but I am confusing myself.
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For answer (iii) are we to solve both probabilities and add them.
1 answer