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A spinner for a board game has the numbers 24, 22, 54, 36, 18, 10, 12 & 64 on it....

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connorcolin3 | (Level 1) Valedictorian

Posted March 21, 2013 at 8:20 PM via web

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A spinner for a board game has the numbers 24, 22, 54, 36, 18, 10, 12 & 64 on it. What is the probability that the spinner will land on a multiple of 3 & 4?

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted March 21, 2013 at 8:51 PM (Answer #1)

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Multiples of 3 and 4 are: 12, 24, 36. All other numbers are not multiples of both 3 and 4 (e.g. 64 is multiple of 4 but not of 3).

So probability that the spinner will land on multiple of 3 and 4 is number of multiples of 3 and 4 divided by number of all numbers.` `


So there are 3 multiples of 3 and 4 out of 8 possible numbers. Hence, the probability of the spinner landing on a multiple of 3 and 4 is

`P=3/8=0.375`

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violy | High School Teacher | (Level 3) Assistant Educator

Posted March 22, 2013 at 12:36 AM (Answer #1)

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We can first first try to prime factorize the given numbers:

24= 2*2*2*3

22` ` = 2* 11

54 = 2 * 3 * 3 * 3

36 = 2* 2* 3 * 3

18 = 2 * 3 * 3

10=   2 * 5

12 = 2 * 2 * 3

64 = 2* 2* 2* 2 *2*2

Hence, the numbers that is both multple of 3 and 4 are 24,36, and 12.

There are 8 number in all, and 3 of them are multiples of 3 and 4.

So, the probability that the spinner will land on a multiple of 3 and 4 is 3/8 or 37.5%.

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted March 23, 2013 at 1:18 PM (Answer #2)

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Lest S be sample space

S={10,12,18,22,24,36,54,64}

E= number multiple of 3 and 4 i.e multiple of 12={12,24,36}

n(S)=8

n(E)=3

P(E)=probability of E=3/8=.375

 

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