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qaws [65]
2 years ago
7

A handbag contains five coins, four keys,

Mathematics
1 answer:
Dmitry [639]2 years ago
5 0

The expected number of mints taken out of the handbag = 0.94

For given question,

We have been given that a handbag contains five coins, four keys,

and eight mints.

Total number of items in the handbag would be,

= 5 coins + 4 keys + 8 mints

= 17 items

So, the total number of items in a handbag = 17

Here, two items are taken out of the handbag one after the other and not replaced.

We need to find the expected number of mints taken out of the handbag.

Number of items taken out = 2

Let x be the number of mints taken out of the handbag.

x can be 0, 1, or 2

For x = 0,

the two items taken out could be either coins or keys.

no. of coins + no. of keys = 9

So, the 1st pick = 9/17

and 2nd pick = 8/16

⇒ P(x = 0) = 9/17 × 8/16

⇒ P(x = 0) = 72/272

For x = 1,

Assume that in the 1st pick mint is taken out.

So, 1st pick = 8/17

For second pick, it can be either coin or key.

This means, there are two possibilities for the second pick.

so, 2nd pick = 2 × 9/16

⇒ P(x = 1) = 8/17 × 9/16 × 2

⇒ P(x = 1) = 144/272

For x = 2,

1st pick = 8/17

2nd pick = 7/16

⇒ P(x = 2) = 8/17 × 7/16

⇒ P(x = 2) = 56/272

So, the expected number of mints taken out of the handbag would be,

= 0 × 72/272 + 1 × 144/272 + 2 × 56/272

= 0.94

Therefore, the expected number of mints taken out of the handbag = 0.94

Learn more about the probability here:

brainly.com/question/3679442

#SPJ4

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(This question is related to the picture question)
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Step-by-step explanation:

==>>We would have to multiply the original amount by 1.50^x because the initial amount would be 1, and 50% increase would be .5 so 1.5 and you raise it to the number of years to show the total increase.

  • Let's test it

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