Answer:
a)
[tex]\sf P(\textsf{gold}) = \dfrac{1}{4} [/tex]
[tex]\sf P(\textsf{silver}) = \dfrac{23}{56} [/tex]
[tex]\sf P(\textsf{bronze}) = \dfrac{19}{56} [/tex]
b) 49 times
Step-by-step explanation:
a) Probability of randomly selecting each kind of coin:
Probability of selecting a gold coin:
[tex]\sf P(\textsf{gold}) = \dfrac{\textsf{no. of gold coins}}{\textsf{total no. of coins}} \\\\= \dfrac{14}{14 + 23 + 19} \\\\ = \dfrac{14}{56}\\\\ = \dfrac{1}{4} [/tex]
Probability of selecting a silver coin:
[tex]\sf P(\textsf{silver}) = \dfrac{\textsf{no. of silver coins}}{\textsf{total no. of coins}} \\\\= \dfrac{23}{14 + 23 + 19}\\\\ = \dfrac{23}{56} [/tex]
Probability of selecting a bronze coin:
[tex]\sf P(\textsf{bronze}) = \dfrac{\textsf{no. of bronze coins}}{\textsf{total no. of coins}}\\\\ = \dfrac{19}{14 + 23 + 19}\\\\ = \dfrac{19}{56} [/tex]
b) To predict the number of times Roger selects a silver coin out of 120 random selections:
[tex]\begin{aligned} \sf \textsf{No. of times selecting a silver coin} & = P(\textsf{silver}) \times \textsf{total no. of selections} \\\\ & = \dfrac{23}{56} \times 120 \\\\ & = \dfrac{23}{56} \times 120 \\\\ & = \dfrac{2760}{56} \\\\ & = 49.285714285714 \\\\ & = 49 \textsf{(in nearest whole number)} \end{aligned} [/tex]
So, we predict that Roger would select a silver coin approximately 49 times out of 120 random selections.
Answer:
[tex]\textsf{A)}\quad \sf P(Gold)=\dfrac{1}{4}, \;\;P(Silver)=\dfrac{23}{56},\;\;P(Bronze)=\dfrac{19}{56}[/tex]
[tex]\textsf{B)} \quad \sf 49\;times[/tex]
Step-by-step explanation:
A bag of coins contains 14 gold coins, 23 silver coins, and 19 bronze coins, which means the total number of coins is:
[tex]\sf Total\;number\;of\;coins=14+23+19=56[/tex]
Probability is a measure of the likelihood of an event occurring and is calculated by dividing the number of favorable outcomes by the total number of possible outcomes:
[tex]\boxed{\sf Probability=\dfrac{\textsf{Number of favorable outcomes}}{\textsf{Total number of possible outcomes}}}[/tex]
To find the probability of randomly selecting each kind of coin, we divide the number of coins of each type by the total number of coins in the bag:
[tex]\sf P(Gold)=\dfrac{14}{56}=\dfrac{14 \times 1}{14 \times 4}=\dfrac{1}{4}[/tex]
[tex]\sf P(Silver)=\dfrac{23}{56}[/tex]
[tex]\sf P(Bronze)=\dfrac{19}{56}[/tex]
To predict how many times Roger will select a silver coin in 120 trials, we can use the formula for the expected value:
[tex]\boxed{\textsf{Expected value} = \textsf{Number of trials} \times \textsf{Probability of success}}[/tex]
In this case, the number of trials is 120 and the probability of success of selecting a silver coin from the bag is 23/56 (from Part A). Therefore:
[tex]\sf \textsf{Expected number of silver coins} = 120 \times \dfrac{23}{56}\\\\\\\textsf{Expected number of silver coins} = \dfrac{2760}{56}\\\\\\\textsf{Expected number of silver coins}=49.2857142...\\\\\\\textsf{Expected number of silver coins}\approx 49[/tex]
Therefore, we can predict that Roger would select a silver coin approximately 49 times in 120 trials.
