# Maths homework

Answers

Probability = (Possibilities with right outcome) / (Total possibilities)Possibilities with right outcome = "How many are 'black or red' "Total possibilities = "How many counters in total"Answer will be a fraction / decimal between 0 (never happens) and 1 (always happens).Hope this helps.

19 September 2017

Total no. of counters=12Prob. of a black counter =3/12Prob. of a red counter=4/12And the probability of a counter being red and black is 4/12+3/12=7/12.

20 September 2017

Knowledge Required Beforehand:1) Whenever you are asked to find the "Probability of an event A
OR Probability of another event B", just add the favorable
outcomes of event A and event B.Using this, in the question given above, the number of favorable outcomes = no. of black counters + no. of red counters = 3 + 4 = 72) One must also find the total number of possible outcomes.For our problem total number of possible outcomes = number of
Black Counters + number of red Counters + number of Blue Counters = 3 + 4
+ 5 = 12.3) P (E) = Number of Favorable Outcome รท Total Number of Possible Outcomes, where: P(E) stands for Probability of the event.Actual Solution:P (Black or Red Counter) = Number of Black Or Red Counter รท Total Number of Possible Outcomes = 7รท12 = 0.58 (approximately).

28 September 2017

Hello, I can help you solve this kind of questions;Total number of counters 12P(BL or R)=P(BL) + P(R)P(BL or R)=3/12 + 4/12P(BL or R)=7/12

11 October 2017

7 out of 12

22 October 2017

There are 3 black counters and 4 red counters and 12 counters of any colour altogether. When you get the probability of choosing either a red or a black the probability is 4+3 = 7/12

22 October 2017

total ways of taking a counter =12 different ways(as there is total 12 balls)total ways of taking a black counter=3total ways of taking a red counter=4total ways of taking a black or red=3+4=7probabilty of (black or red)=7/12

22 October 2017

3+4+5 = 12 (this is the total amount of counters)3+4= 7 (total of red and black counters)7/12 (this is the simplest fraction of counters that are red or black)

24 October 2017

Firstly, you would need to calculate the total number of counters in the bag. There are 3 black, 4 red and 5 blue. To calculate the total number of counters,we add this all together: 3+4+5=12, so the total number of counters in the bag is 12. The key aspect here is that it asks you to calculate the probability that the counter is black OR red. This means you must find the number of black and red counters in the bag. In this case, it's 7 (3+4). You then divide this number (n) by the total number of counters (#). So probability (P) = n/#. So P = 7/12.

25 October 2017

Total number of counters = 3+4+5=12probability of picking a black counter, P(black) = 3/12probability of picking a red counter, P(red) = 4/12probability of picking a blue counter, P(blue) = 5/12'and' means multiply the probabilities stated'or' means add the probabilities statedso probability that a black or red counter is taken = P(black) + P(red)= 3/12 + 4/12 = 7/12

29 October 2017

3 black + 4 red = 7 counters that are black or redTotal number of counters = 3+4+5= 12 countersprobability = outcome we want / total possible outcomesprob = 7/12 chance of a black or red counter

05 November 2017

solution.docx

07 November 2017

14 March 2018

New_Doc_2018-03-24__2__1_3_.jpg

24 March 2018

P(Red) = 4/12 = 1/3P(Black) = 3/12 = 1/3When i say P(Black) i mean what is the probability that the counter is black

09 April 2018

number of counters in the bag = 3 + 4 + 5 = 12so the probability of getting red counter equals the number of red counters in bag divided by the total number of countersP(red) = 4 / 12 = 1/3and we can do the same for P(black)P(black) = 3 / 12 = 1 / 4

01 July 2018

Altogether there are 12 counters the probability of a black counter is 3/12the probability of a red counter is 4/12 Because the questions asks for a black OR red counterwe have to apply the or real which states you add the probabilities rather than times which is the and rule therefore 3/12 + 4/12 = 7/12

01 July 2018

3/12+4/12=7/12

07 October 2018

Total sample space= (3 black + 4 red + 5 blue) counters=12 countersprobability of drawing a black counter= 3/12probability of drawing a red counter=4/12probability of drawing a black or red counter= (3+4)/12=7/12

22 December 2018

no. of favourable outcomes = 3+4 =7total outcomes = 3+4+5 =12Thus, probability is 7/12

02 October 2019

12 October 2019

12 counters totalP(black) = 3/12P(red) = 4/12P(black or red) = (3+4)/12 = 7/12

23 January 2020

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