7 black shirts 5 white shirts 10 gray shirts one is chosen at random, what is the probability that it is not gray [U]Find the total shirts:[/U] Total shirts = Black Shirts + White Shirts + Gray Shirts Total shirts = 7 + 5 + 10 Total shirts = 22 [U]Calculate the probability of choosing a gray shirt:[/U] P(Gray) = Number of Gray shirts / Total Shirts P(Gray) = 10/22 We can simplify this fraction. We [URL='https://www.mathcelebrity.com/fraction.php?frac1=10%2F22&frac2=3%2F8&pl=Simplify']type in 10/22 into our search engine, choose simplify[/URL], and we get: P(Gray) = [B]5/11[/B]

A bag contains 120 marbles. Some are red and the rest are black. There are 19 red marbles for every black marble. How many red marbles are in the bag? Let the red marbles be r Let the black marbles be b. A 19 to 1 red to black is written as: r = 19b We're also given: b + r = 120 Substitute r = 19b into this equation and we get: b + 19b = 120 Combine like terms: 20b = 120 To solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=20b%3D120&pl=Solve']we type it in our search engine [/URL]and we get: b = 6 Since r = 19b, we substitute b = 6 into this equation to solve for r: r = 19(6) r = [B]114[/B]

A bag contains 120 marbles. Some are red and the rest are black. There are 19 red marbles for every black marble. How many red marbles are in the bag? Using our [URL='https://www.mathcelebrity.com/ratio.php?simpratio=100%3A350&rs=19%3A1&rtot=120&pl=Calculate+Ratio']ratio calculator[/URL], we get: [LIST] [*]Red = 114 [*]Black = 6 [/LIST]

A bag contains 3 black, 4 red, 3 yellow, and 2 green marbles. What is the probability of drawing a black and then a red marble out of the bag without replacing the black marble before drawing the red marble? The phrase [U][B]without replacement[/B][/U] is a huge clue on this problem. Take each draw and calculate the probability. Draw 1: P(Drawing a red) P(Drawing a red) = Total Red marbles n the jar / Total marbles in the jar P(Drawing a red) = 4/12 4/12 simplifies to 1/3 using a common factor of 4: P(Drawing a red) = 1/3 Draw 2: P(Drawing a black) P(Drawing a black) = Total Black marbles in the jar / Total marbles in the jar [I]We drew one red marble already. Without replacement means we do not put it back. Therefore, we have 12 - 1 = 11 marbles left in the jar.[/I] P(Drawing a black) = 3/11 The question asks, what is the the following probability: P(Drawing a Red, Drawing a Black) Because each draw is [I][U]independent[/U], [/I]we multiply each draw probability together: P(Drawing a Red, Black) = P(Drawing a Red) * P(Drawing a Black) P(Drawing a Red, Black) = 1/3 * 3/11 P(Drawing a Red, Black) = [B]1/11[/B]

A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked random. Then probability of that the ball will be white is: Probability that you pick any bag is 0.5. Bag 1 White Ball = 0.5(3/5) = 3/10 = 0.3 Bag 2 White Ball = 0.5(2/6) = 1/6 = 0.16667 Add them both: 0.3 + 0.16667 = [B]0.46667[/B]

A box contains 5 black and 2 white balls. 2 balls are drawn without replacement. Find the probability of drawing 2 black balls. First draw probability of black is: Total Balls in box = Black balls + white balls Total Balls in Box = 5 + 2 Total Balls in Box = 7 P(Black) = Black Balls / Total balls in box P(Black) = 5/7 Second draw probability of black (with no replacement) is: Total Balls in box = Black balls + white balls Total Balls in Box = 4 + 2 Total Balls in Box = 6 P(Black) = Black Balls / Total balls in box P(Black) = 4/6 Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F6&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL], we see that 4/6 is: 2/3 Since each event is independent, we can multiply them to find the probability of drawing 2 black balls: P(Black, Black) = 5/7 * 2/3 [URL='https://www.mathcelebrity.com/fraction.php?frac1=5%2F7&frac2=2%2F3&pl=Multiply']P(Black, Black)[/URL] = 10/21 [MEDIA=youtube]HEa_G3nwgUQ[/MEDIA]

A card is drawn from a pack of 52 cards. The probability that the card drawn is a red card is The deck is split evenly between red and black cards. So we have 52/2 = 26 red cards P(Red) = # of Red Cards / Total Deck Cards P(Red) = 26/52 We can simplify this fraction. [URL='https://www.mathcelebrity.com/fraction.php?frac1=26%2F52&frac2=3%2F8&pl=Simplify']Using our fraction calculator[/URL], we get: P(Red) = [B]1/2 or 0.5[/B]

A card is picked from a deck of 52 cards. Find the probability of getting a black ace or a red queen. In a standard deck of 52 cards, we have: [LIST] [*]2 black Aces with probability 2/52 = 1/26 [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F52&frac2=3%2F8&pl=Simplify']using our fraction simplifier[/URL] [*]2 red Queens with probability 2/52 = 1/26 [URL='http://using our fraction simplifier']using our fraction simplifier[/URL] [/LIST] The problems asks for P(Red Queen Or Black Ace). Or means we add, so we have: P(Red Queen Or Black Ace) = P(Red Queen) + P(Black Ace) P(Red Queen Or P Black Ace) = 1/26 + 1/26 P(Red Queen Or P Black Ace) = 2/26 P(Red Queen Or P Black Ace) = [B]1/13[/B] [URL='https://www.mathcelebrity.com/fraction.php?frac1=2%2F26&frac2=3%2F8&pl=Simplify']using our fraction simplifier[/URL]

a card selected from a deck of 52 cards what is the probability it is a black card or face card Facts: [LIST] [*]Half the cards in the deck are black (26/52) [*]There are 12 face cards (K, Q, J) in a deck (12/52) [*]Black and Face = 6/52 (Duplicates from above) [/LIST] P(Black or Face) = P(Black) + P(Face) - P(Black And Face) P(Black or Face) = 26/52 + 12/52 - 6/52 P(Black or Face) = 32/52 We can simplify this. We use our [URL='https://www.mathcelebrity.com/fraction.php?frac1=32%2F52&frac2=3%2F8&pl=Simplify']fraction simplifier[/URL] to get: P(Black or Face) = [B]8/13[/B]

A drawer is filled with 9 black shirts , 6 white shirts, and 5 gray shirts one shirt is chosen at random from the drawer find the probability that it is not a white shirt P(Not White) = P(Black or Gray) P(Black or Gray) = (Total Black + Total Gray)/Total Shirts P(Black or Gray) = (9 + 5)/(9 + 6 + 5) P(Black or Gray) = 14/20 Simplifying this [URL='https://www.mathcelebrity.com/fraction.php?frac1=14%2F20&frac2=3%2F8&pl=Simplify']using our fraction simplify calculator[/URL], we get: P(Black or Gray) = [B]7/10, or 0.7 or 70%[/B]

A laundry basket contains 30 socks, of which 9 are black. What is the probability that a randomly selected sock will be black? P(Black) = 9/30 Simplifying, we can divide top and bottom by 3: [B]3/10 3/10 as a percentage is 30%[/B]

A quarter of the learners in a class have blond hair and two thirds have brown hair. The rest of the learners in the class have black hair. How many learners in the class if 9 of them have blonde hair? Total learners = Blond + Brown + Black Total Learners = 1/4 + 2/3 + Black Total Learners will be 1, the sum of all fractions 1/4 + 2/3 + Black = 1 Using common denominators of 12, we have: 3/12 + 8/12 + Black = 12/12 11/12 + Black = 12/12 Subtract 11/12 from each side: Black = 1/12 Let t be the total number of people in class. We are given for blondes: 1/4t = 9 Multiply each side by 4 [B]t = 36[/B] Brown Hair 2/3(36) = 24 Black Hair 1/12(36) = 3

A spinner has 3 equal sections labelled A, B, C. A bag contains 3 marbles: 1 grey, 1 black, and 1 white. The pointer is spun and a marble is picked at random. a) Use a tree diagram to list the possible outcomes. [LIST=1] [*][B]A, Grey[/B] [*][B]A, Black[/B] [*][B]A, White[/B] [*][B]B, Grey[/B] [*][B]B, Black[/B] [*][B]B, White[/B] [*][B]C, Grey[/B] [*][B]C, Black[/B] [*][B]C, White[/B] [/LIST] b) What is the probability of: i) spinning A? P(A) = Number of A sections on spinner / Total Sections P(A) = [B]1/3[/B] --------------------------------- ii) picking a grey marble? P(A) = Number of grey marbles / Total Marbles P(A) = [B]1/3[/B] --------------------------------- iii) spinning A and picking a white marble? Since they're independent events, we multiply to get: P(A AND White) = P(A) * P(White) P(A) was found in i) as 1/3 Find P(White): P(White) = Number of white marbles / Total Marbles P(White) = 1/3 [B][/B] Therefore, we have: P(A AND White) = 1/3 * 1/3 P(A AND White) = [B]1/9[/B] --------------------------------- iv) spinning C and picking a pink marble? Since they're independent events, we multiply to get: P(C AND Pink) = P(C) * P(Pink) Find P(C): P(C) = Number of C sections on spinner / Total Sections P(C) = 1/3 [B][/B] Find P(Pink): P(Pink) = Number of pink marbles / Total Marbles P(Pink) = 0/3 [B][/B] Therefore, we have: P(C AND Pink) = 1/3 * 0 P(C AND Pink) = [B]0[/B]

A store manager must calculate the total number of winter hats available to sell in the store from a starting number of 293. In the past month, the store sold 43 blue hats, 96 black hats, 28 red hats, and 61 pink hats. The store received a shipment of 48 blue hats, 60 black hats, 18 red hats, and 24 pink hats. How many total hats does the store have for sale? [LIST=1] [*]We start with 293 hats [*]We calculate the hats sold: (43 + 96 + 28 + 61) = 228 [*]We subtract Step 2 from Step 1 to get remaining hats before the shipment: 293 - 228 = 65 [*]Now we calculate the number of hats received in the shipment: (48 + 60 + 18 + 24) = 150 [*]We add Step 4 to Step 3: 65 + 150 = [B]215 hats for sale[/B] [/LIST]

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for every 10 white cars a dealer sells he sells 7 silver, 6 blue, 5 red, 4 yellow, 3 green, 2 black, 2 purple and 1 brown car. If he sells 120 cars how many blue cars? [U]Take this in blocks, so each block has:[/U] 10 white + 7 silver + 6 blue + 5 red + 4 yellow + 3 green + 2 black + 2 purple + 1 brown = 40 cars [U]Calculate the number of blocks:[/U] 120 cars / 40 cars = 3 blocks. [U]For 120 cars sold, it takes 3 blocks, which means we multiply:[/U] 6 blue cars per block * 3 blocks = [B]18 blue cars[/B]

Gregg has 8 cards.Half red,half black. He picks 2 cards from the deck.What is the probability both of them are red? Half means 4 cards are red and 4 cards are black. The first draw probability of red is: 4 total red cards out of 8 total cards = 4/8. [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F8&frac2=3%2F8&pl=Simplify']Simplified, this is[/URL] 1/2 The second draw is 3 total red cards out of 7 remaining cards. Since 1 red was drawn (4 - 1) = 3 reds left and 1 card was drawn (8 -1) = left 3/7 Since each draw is independent, we multiply the probabilities: 1/2 * 3/7 = [B]3/14[/B]

if the blackout begins at 5:20 pm and ended at 7:05 pm how long did the black out last? [I]add[/I] 2 hours, and we get: 7:20 [I]Subtract[/I] 15 minutes, and we get: 7:05 2 hours - 15 minutes = [B]1 hour and 45 minutes[/B]

In a certain lot, there are 16 white, 7 red, 8 blue, and 9 black cars. You randomly pick a set of keys to one of the cars. What is the probability of choosing a set of keys to a blue car? [U]Our total cars are:[/U] Total Cars = White Cars + Red Cars + Blue Cars = Black Cars Total Cars = 16 + 7 + 8 + 9 Total Cars = 40 P(Blue) = Blue Cars / Total Cars P(Blue) = 8/40 Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=8%2F40&frac2=3%2F8&pl=Simplify']fraction simplify calculator[/URL], we get: P(Blue) = [B]1/5[/B]

On your first draw, what is the probability of drawing a red card, without looking, from a shuffled deck containing 6 red cards, 6 blue cards, and 8 black cards? P(Red) = Total Red / Total Cards P(Red) = 6 red/(6 red + 6 blue + 8 black) P(Red) = 6/20 This fraction can be simplified. The [URL='https://www.mathcelebrity.com/gcflcm.php?num1=6&num2=20&num3=&pl=GCF+and+LCM']greatest common factor of 6 and 20[/URL] is 2. So we divide top and bottom of our probability by 2: P(Red) = 6/2 / 20 / 2 P(Red) = [B]3/10[/B]

She earns $20 per hour as a carpenter and $25 per hour as a blacksmith, last week Giselle worked both jobs for a total of 30 hours, and a total of $690. How long did Giselle work as a carpenter and how long did she work as a blacksmith? Assumptions: [LIST] [*]Let b be the number of hours Giselle worked as a blacksmith [*]Let c be the number of hours Giselle worked as a carpenter [/LIST] Givens: [LIST=1] [*]b + c = 30 [*]25b + 20c = 690 [/LIST] Rearrange equation (1) to solve for b by subtracting c from each side: [LIST=1] [*]b = 30 - c [*]25b + 20c = 690 [/LIST] Substitute equation (1) into equation (2) for b 25(30 - c) + 20c = 690 Multiply through: 750 - 25c + 20c = 690 To solve for c, we [URL='https://www.mathcelebrity.com/1unk.php?num=750-25c%2B20c%3D690&pl=Solve']type this equation into our search engine[/URL] and we get: c = [B]12 [/B] Now, we plug in c = 12 into modified equation (1) to solve for b: b = 30 - 12 b = [B]18[/B]

The Henson family cleaned out all their drawers. They found 47 black pens and 39 blue pens. They also found 6 pens in other colors. How many pens did they find in all? The phrase [I]in all[/I] means we add, so we have: Total pens = Black Pens + Blue Pens + Other color pens Total pens = 47 + 39 + 6 Total pens = [B]92[/B]

There are 5 orange books, 12 black books, and 8 tan books on Mr. Johnsons bookshelf. Calculate the probability of randomly selecting a black book and then a tan book without replacement. Write your answer as a percent. P(black book first draw) P(black book first draw) = 12 black / (5 orange + 12 black + 8 tan) P(black book first draw) = 12 / 25 P(tan book second draw) P(tan book second draw) = 8 tan / (5 orange + 11 black + 8 tan) <-- 11 black because we already drew one black P(tan book second draw) = 8 / 24 Using our fraction reduction calculator, this simplifies to 1/3 Since each draw is independent, we multiply both probabilities: P(black book first draw, tan book second draw) = 12/25 * 1/3 P(black book first draw, tan book second draw) = 12/75 P(black book first draw, tan book second draw) = [B]16%[/B]

There are 5 red and 4 black balls in a box. If you pick out 2 balls without replacement, what is the probability of getiing at least one red ball? First list out our sample space. At least one means 1 or 2 red balls, so we have 3 possible draws: [LIST=1] [*]Red, Black [*]Black, Red [*]Red, Red [/LIST] List out the probabilities: [LIST=1] [*]Red (5/9) * Black (4/8) = 5/18 [*]Black (4/9) * Red (5/8) = 5/18 [*]Red (5/9) * Red (4/8) = 5/18 [/LIST] Add these up: 3(5)/18 = [B]5/6[/B]

There are 50 pairs of pants. One-half of the pants are black. One-fifth of the pants are tan. How many pairs of pants are not black or tan. First, determine what fraction of pants are black and tan: 1/2 + 1/5 Using our [URL='https://www.mathcelebrity.com/fraction.php?frac1=1%2F2&frac2=1%2F5&pl=Add']fraction addition calculator[/URL], we get 7/10. So the rest of the pants are 1 - 7/10. 1 can be written as 10/10. So we have 10/10 - 7/10 = 3/10 3/10 * 50 = 150/10 = [B]15[/B]

you draw a card at random from a deck that contains 3 black cards and 7 red cards what is the probability of you drawing a black card Total cards = 3 black + 7 red Total cards = 10 P(Black) = Black cards / Total Cards P(Black) = [B]3/10 or 0.3[/B]