Question

Help me please..30points!!!!

Answer 1

Explanation:

a) 10/18 · 9/17

5/9 · 9/17

5/17

b) 8/18 · 7/17

4/9 · 7/17

28/153

c) 10/18 · 8/17

5/9 · 8/17

40/153

Answer 2

`\textsf{a)} \quad (5)/(17)=29.4\%`

`\textsf{b)} \quad (28)/(153)=18.3\%`

`\textsf{c)} \quad (40)/(153)=26.1\%`

Explanation:

Bag of 18 fruit drop sweets:

• 10 apple flavoured
• 8 blackberry flavoured

`\boxed{\sf Probability\:of\:an\:event\:occurring = (Number\:of\:ways\:it\:can\:occur)/(Total\:number\:of\:possible\:outcomes)}`

Part (a)

The probability of the first sweet chosen being apple flavoured is:

`\implies \sf P(apple)=(10)/(18)`

Now there are 9 apple flavoured sweets and a total of 17 sweets remaining, so the probability of choosing a second apple flavoured sweet is:

`\implies \sf P(apple)=(9)/(17)`

Therefore, the probability of both sweets being apple flavoured is:

`\implies \sf P(apple)\;and\;P(apple)= (10)/(18) * (9)/(17)=(5)/(17)`

Part (b)

The probability of the first sweet chosen being blackberry flavoured is:

`\implies \sf P(blackberry )=(8)/(18)`

Now there are 7 blackberry flavoured sweets and a total of 17 sweets remaining, so the probability of choosing a second blackberry flavoured sweet is:

`\implies \sf P(blackberry )=(7)/(17)`

Therefore, the probability of both sweets being blackberry flavoured is:

`\implies \sf P(blackberry )\;and\;P(blackberry )= (8)/(18) * (7)/(17)=(28)/(153)`

Part (c)

The probability of the first sweet chosen being apple flavoured is:

`\implies \sf P(apple)=(10)/(18)`

Now there are 9 apple flavoured sweets, 8 blackberry sweets, and a total of 17 sweets remaining, so the probability of choosing a blackberry flavoured sweet is:

`\implies \sf P(blackberry)=(8)/(17)`

Therefore, the probability of the first sweet being apple flavoured and the second sweet being blackberry flavoured is:

`\implies \sf P(apple)\;and\;P(blackberry)= (10)/(18) * (8)/(17)=(40)/(153)`