Kyle has three short straws, four medium straws, and six long straws. If he randomly draws two straws, one at a time without replacement, what is the probability that both are short straws? A) 1/26 B) 3/52 C) 6/169 D) 9/169

lets call these SSS MMMM LLL LLL for short , medium and long

so the first straw, you have a probability of 3 / 13 of picking a short straw. the next straw, now you only have 12 straws since you already picked a short straw from last choice. so now you have 2/ 12 of picking a short straw so 3/13 * 2 / 12

more generally, the probability of the conjunction (AND) of two events is P ( A & B ) = P ( A) * P ( B | A) , P(B|A) means the probability of given that A has occurred already. So here we have probability ( short straw on first draw & short straw on second draw) = P(straw on first draw) * P(straw on second draw given first straw was a short one)

i just want the answer. this is confusing enough

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