green banana

Probability Level 1

Marvin roll a die. If he obtain a $5$ , he chooses a banana from box $1$ where $4$ are green and $6$ are yellow. If he obtain a number that is not a $5$ , he chooses a banana from box $2$ where $8$ bananas are green and $3$ bananas are yellow. What is the probability of obtaining a colored green banana? Your answer can be expressed as $\dfrac{M}{K}$ where $M$ and $K$ are positive co-prime integers. Find $M+K$ .

The answer is 92.

2 solutions

Darsh Kedia
May 15, 2020

The probability of getting a 5 on a dice is $\frac{1}{6}$ .

$\therefore$ The probability of getting a green banana after getting a 5 on the dice is $\frac { 1 }{ 6 } \times \frac { 4 }{ 4+6 } =\frac { 1 }{ 6 } \times \frac { 4 }{ 10 } =\frac { 1 }{ 15 }$

The probability of getting a number other than 5 on a dice is $\frac{5}{6}$ .

$\therefore$ The probability of getting a green banana after getting a number other than 5 is $\frac { 5 }{ 6 } \times \frac { 8 }{ 8+3 } =\frac { 5 }{ 6 } \times \frac { 8 }{ 11 } =\frac { 20 }{ 33 }$

The total probability of getting a green banana is the sum of the individual probabilities.

$\therefore$ The Probability Of Getting A Green Banana = $\frac { 1 }{ 15 } +\frac { 20 }{ 33 } =\frac { 100+11 }{ 165 } =\frac { 111 }{ 165 } =\frac { 37 }{ 55 }$

The Required Answer Is: $\Large M+K=37+55=\boxed { 92 }$

Marvin Kalngan
May 15, 2020

$P_{green}=P(G\times B_1 \cup G\times B_2)=P(G\times B_1)+P(G\times B_2)=\dfrac{1}{6}\left(\dfrac{4}{10}\right)+\dfrac{5}{6}\left(\dfrac{8}{11}\right)=\dfrac{37}{55}$

The required answer is: $M+K=37+55=\boxed{92}$

green banana

The answer is 92.

2 solutions

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