Grade 7 · Statistics & probability

Two-event conditional probability word problems practice

Two-event conditional probability word problems is a grade 7 math skill aligned to Common Core standard 7.SP.C.8: find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 two-event conditional probability word problems problems our math games drill.

CCSS 7.SP.C.810 questions in the bank
Sample questions

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Question 1easy

In a class of 2020 students, 1212 students bring lunch from home, and 44 of those students also bring a reusable water bottle. If a student who brings lunch from home is selected at random, what is the probability that the student also brings a reusable water bottle?

Restrict to the 1212 students who bring lunch from home. 44 of them also bring a bottle, so P(bottlelunch from home)=412=13P(\text{bottle} \mid \text{lunch from home}) = \dfrac{4}{12} = \dfrac{1}{3}.

Question 2easy

The letters of the word BANANA are placed in a bag. Two letters are selected one at a time without replacement. What is the probability of drawing an A and then an N?

The word BANANA has 66 letters with 33 A's and 22 N's. P(A first)=36P(\text{A first}) = \dfrac{3}{6} and P(N secondA first)=25P(\text{N second} \mid \text{A first}) = \dfrac{2}{5}, so P(A then N)=36×25=15P(\text{A then N}) = \dfrac{3}{6} \times \dfrac{2}{5} = \dfrac{1}{5}.

Question 3easy

The letters of the word STRAWBERRY are placed in a bag. Two letters are selected one at a time without replacement. What is the probability of drawing an R and then another R?

The word STRAWBERRY has 1010 letters with 33 R's. P(R first)=310P(\text{R first}) = \dfrac{3}{10} and P(R secondR first)=29P(\text{R second} \mid \text{R first}) = \dfrac{2}{9}, so P(R then R)=310×29=115P(\text{R then R}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{1}{15}.

Question 4medium

The letters of the word TOMORROW are placed in a bag. Two letters are selected one at a time without replacement. What is the probability of drawing an R and then a T?

The word TOMORROW has 88 letters with 22 R's and 11 T. P(R first)=28P(\text{R first}) = \dfrac{2}{8} and P(T secondR first)=17P(\text{T second} \mid \text{R first}) = \dfrac{1}{7}, so P(R then T)=28×17=128P(\text{R then T}) = \dfrac{2}{8} \times \dfrac{1}{7} = \dfrac{1}{28}.

Question 5medium

The letters of the word COLOR are placed in a bag. Two letters are selected one at a time without replacement. What is the probability of drawing an O and then another O?

The word COLOR has 55 letters with 22 O's. P(O first)=25P(\text{O first}) = \dfrac{2}{5} and P(O secondO first)=14P(\text{O second} \mid \text{O first}) = \dfrac{1}{4}, so P(O then O)=25×14=110P(\text{O then O}) = \dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{1}{10}.

Question 6medium

The letters of the word MISSISSIPPI are placed in a bag. Two letters are selected one at a time without replacement. What is the probability of drawing an S and then another S?

The word MISSISSIPPI has 1111 letters with 44 S's. P(S first)=411P(\text{S first}) = \dfrac{4}{11} and P(S secondS first)=310P(\text{S second} \mid \text{S first}) = \dfrac{3}{10}, so P(S then S)=411×310=655P(\text{S then S}) = \dfrac{4}{11} \times \dfrac{3}{10} = \dfrac{6}{55}.

Question 7medium

A bag contains 66 red marbles and 44 white marbles. Two marbles are drawn one at a time without replacement. The first marble drawn is red. What is the probability that the second marble is white?

After one red marble is removed, 44 white marbles remain out of 99 marbles in the bag. So P(second whitefirst red)=49P(\text{second white} \mid \text{first red}) = \dfrac{4}{9}.

Question 8medium

A bag contains 44 black marbles and 66 white marbles. Two marbles are drawn one at a time without replacement. The first marble drawn is black. What is the probability that the second marble is also black?

After one black marble is removed, 33 black marbles remain out of 99 marbles in the bag. So P(second blackfirst black)=39=13P(\text{second black} \mid \text{first black}) = \dfrac{3}{9} = \dfrac{1}{3}.

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