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3 years ago

Two fair dice are rolled. what is the probability that they sum to: even

2 answers:

4 0

Total number of possible outcomes = 36

Total number of outcomes with sum even = 18

P(even) = 18/36 = 1/2

Answer: 1/2

sladkih [1.3K]3 years ago

4 0

$|\Omega|=6^2=36\\ |A|=\underbrace{3\cdot3}_{\text{O,O}}+\underbrace{3\cdot3}_{\text{E,E}}=18\\\\ P(A)=\dfrac{18}{36}=\dfrac{1}{2}=50\%$

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3 years ago

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jake added 1/8 cup of sunflower seeds and 4/5 cup of banana chips to his sundae. which is the best estimate of the total amount

bagirrra123 [75]

Jake added $\frac{37}{40}$ cup of toppings to his sundae.

Step-by-step explanation:

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Learn more about fractions at:

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2 years ago

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emmainna [20.7K]

Answer:

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2 years ago

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2 years ago

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec

Sphinxa [80]

<span>1.

P(</span>at least one of the calculators is defective)=

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

=n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways,

where $C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360$

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

$C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825$

4. so, P(none of the selected calculators is defective)= $\frac{52,360}{292,825} =0.18$

5. P(at least one of the calculators is defective)=

1- P(none of the selected calculators is defective)=1-0.18=0.82

Answer:0.82

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3 years ago