Question

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.
Group of answer choices

0.11500044682

0.86898036859

0.01601918459

0.13101963141

Answer 1

*100% Correct answers

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

Answer 2

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 = $\frac{5}{6}$

probability of not getting a 5 any of the 4 dies is calculated by multiplying $\frac{5}{6}$ four times. So we get

$=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}$

$=\frac{625}{1296}$

Now, probability of getting a 5 on single die is = $\frac{1}{6}$

Probability of getting a 5 any one of the 4 dies = $\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}$

We have, four dies, so we can get 5 on any of them, hence we need to multiply with 4. So we get

$= \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times 4$

$=\frac{500}{1296}$

Now, probability of atleast 2 out of dices not landing on 5 = $\frac{625}{1296} + \frac{500}{1296}$

$= \frac{1125}{1296} = 0.8680$

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