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Question 1
A surgical technique is performed on 10 patients. You are told there is an 80% chance of success. Find the probability that the surgery is successful for exactly 6 patients.
0.088
Question 2
Sam creates a game in which the player rolls 4 dice. What is the probability in this game of having at least 2 of the dice land on 5.
0.13101963141
Question 3
Katie creates a game in which 3 dimes are flipped at the same time.
5 points are awarded if all 3 dimes land on tails, but no points are awarded for anything else. What is the probability of not getting any points? (Hint: Find the complement).
0.875
Option D
ANSWER:
Sam creates a game in which the player rolls 4 dice .The probability in this game of having at least 2 of the dice land on 5 is 0.1310196314
SOLUTION:
We need to find the probability of atleast 2 out of 4 dices landing on 5.
For our convenience, we will calculate the probability of atleast 2 out of dices not landing on 5, later we can subtract it form 1, so that we will get the required answer.
We know that P(Event) + P (not an Event) = 1 ------------ eqn (1)
There is only one “5” on the die out of six faces. So probability of not getting a 5 on single die is = [tex]\frac{5}{6}[/tex]
probability of not getting a 5 any of the 4 dies is calculated by multiplying [tex]\frac{5}{6}[/tex] four times. So we get
[tex]=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}[/tex]
[tex]=\frac{625}{1296}[/tex]
Now, probability of getting a 5 on single die is = [tex]\frac{1}{6}[/tex]
Probability of getting a 5 any one of the 4 dies = [tex]\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}[/tex]
We have, four dies, so we can get 5 on any of them, hence we need to multiply with 4. So we get
[tex]= \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times 4[/tex]
[tex]=\frac{500}{1296}[/tex]
Now, probability of atleast 2 out of dices not landing on 5 = [tex]\frac{625}{1296} + \frac{500}{1296}[/tex]
[tex]= \frac{1125}{1296} = 0.8680[/tex]
Using equation (1) = P(Event) + P(not an event) = 1
P(Event) + 0.8680 = 1
P(Event) = 1 – 0.8680
P(Event) = 0.1320
By observing the options we can conclude that answer is 0.1310196314
