Complete the following statements based on the pattern you have observed, without a tree diagram. a) If a coin was tossed 5 times, the probability of getting 5 heads would be: ______________ b) If a coin was tossed 6 times, the probability of getting 6 heads would be: ______________c) If a coin was tossed 7 times, the probability of getting 7 heads would be: ______________ d) If a coin was tossed 10 times, the probability of getting 10 heads would be: ______________ 16. Explain how you used the pattern to answer question 15 a,b,c,d 17. Justify (explain) why your rule works for any number of times a coin is tossed
Question
Complete the following statements based on the pattern you have observed, without a tree diagram. a) If a coin was tossed 5 times, the probability of getting 5 heads would be: ______________ b) If a coin was tossed 6 times, the probability of getting 6 heads would be: ______________c) If a coin was tossed 7 times, the probability of getting 7 heads would be: ______________ d) If a coin was tossed 10 times, the probability of getting 10 heads would be: ______________ 16. Explain how you used the pattern to answer question 15 a,b,c,d 17. Justify (explain) why your rule works for any number of times a coin is tossed
Solution
a) If a coin was tossed 5 times, the probability of getting 5 heads would be: 1/32 b) If a coin was tossed 6 times, the probability of getting 6 heads would be: 1/64 c) If a coin was tossed 7 times, the probability of getting 7 heads would be: 1/128 d) If a coin was tossed 10 times, the probability of getting 10 heads would be: 1/1024
-
The pattern used to answer question 15 a,b,c,d is based on the principle that each coin toss is an independent event with two possible outcomes - heads or tails. Therefore, the probability of getting a head in each toss is 1/2. To find the probability of getting heads in all tosses, we multiply the probabilities of each individual event. Hence, for 5 tosses, the probability is (1/2)^5 = 1/32, for 6 tosses, it's (1/2)^6 = 1/64, for 7 tosses, it's (1/2)^7 = 1/128, and for 10 tosses, it's (1/2)^10 = 1/1024.
-
The rule works for any number of times a coin is tossed because each toss is an independent event with two possible outcomes. The probability of getting a head in each toss is always 1/2, regardless of how many times the coin is tossed. Therefore, the probability of getting heads in all tosses is always (1/2) to the power of the number of tosses.
Similar Questions
Make a tree diagram that shows the sample space of rolling a cube with faces numbered 1 through 6 and flipping a fair coin. Using this tree diagram, what is the probability of rolling the number 5 and the coin landing heads up?
TOSSING COINS Create a tree diagram to show the sample space of tossing a coin 4 times then answer the questions below:Q 1.1Level 1- 2 (i) You toss a coin one time 1. List the sample space __________ 2. The number of events in the sample space is _________ 3. The probability of getting 1 head is __________
An experiment consists of tossing three unbiased coins simultaneously. Drawing a probability tree for this experiment will show that the number of events in this experiment is:Group of answer choicesA. 3.B. 6.C. 9.D. None of these choices are correct.
A spinner with 6 possibilities and a coin flip are combined in a tree diagram. How many outcomes are possible?
This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing 7 red (R), 4 yellow (Y) and 5 blue (B) beads. For the coin, P(H) = 13 and P(T) = 23 where H is heads and T is tails.Find P(tossing a Head on the coin AND a Red bead). (Enter your answer as a fraction.)
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.