There are 7 balls in total.
4 are red and 1 is the blue "5" ball. Thus there are five balls out of the seven which meet the criteria and therefore the odds must be 5/7.
The probabilities are mutually exclusive (i.e. you can not both draw a red ball and the "5" ball at the same time as the "5" is blue) so you just add the probabilities of each together.
Probability of drawing a red ball = 4/7 (4 of the 7 balls are red)
Probability of drawing the "5" ball = 1/7 (1 of the 7 balls is the "5" ball)
Overall probability = 4/7 + 1/7 = 5/7
It is 0.6050
The probability that you pick 3 white balls is 0.05 or 5%.----------------------------------------------------------------------------------------------------EXPLANATIONThe probability of first drawing a white ball is: P(W1) =3/6.The probability of drawing a white ball given the event that you already draw awhite ball in the first draw and not replacing it back is: P(W2│W1) =2/5.The probability of drawing a white ball on the third draw given the event that awhite ball was drawn in the first and in the second draw is: P(W3│(W1UW2)) =1/4Now, the probability of drawing 3 white balls one by one with out replacement(taking all 3 balls at a time gets the same analysis and result) is:P(W1UW2UW3) =P(W1)∙P(W2│W1)∙P(W3│(W1UW2)) =(3/6)∙(2/5)∙(1/4) =1/20=0.05 =5%.
Yes, it certainly can if there is only one possible outcome. For instance, the probability of drawing a red ball from a bag containing nothing but red balls is equal to one.
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
25.1%The probability of having at least two same number balls when 7 balls are drawn withreplacement from a box containing 75 numbered balls from 1 to 75 is calculated asfollows:We calculate the probability of drawing 7 times a ball (with replacement) and havingall different numbers. Then we subtract this value from 1. This will be the probabilityof all the possibilities of combinations of 7 numbers where at least two are repeated.1.- For the 1st ball the options are 75/75.2.- For the 2nd ball the options for not repeating the 1st number are 74/75.3.- For the 3rd ball the options for not repeating the previous two numbers are 73/75.4.- For the 4th ball the options for not repeating the previous three numbers 72/75.5.- For the 5th ball ........................................................................... 71/75.6.- For the 6th ball ........................................................................... 70/75.7.- For the 7th ball ........................................................................... 69/75.The probability of drawing seven balls (with replacement) having all different numbersis: 75/75(74/75)(73/75)(72/75)(71/75)(70/75)(69/75) = 0.749419476...The probability of having at least two same number balls when 7 balls are drawn with replacement from the given box is:P = 1 - 0.749419476... = 0.250580524...≈ 25.1%
It is 0.6050
The probability is (6/14)*(5/13) = 30/182 = 0.1648 approx.
The probability that you pick 3 white balls is 0.05 or 5%.----------------------------------------------------------------------------------------------------EXPLANATIONThe probability of first drawing a white ball is: P(W1) =3/6.The probability of drawing a white ball given the event that you already draw awhite ball in the first draw and not replacing it back is: P(W2│W1) =2/5.The probability of drawing a white ball on the third draw given the event that awhite ball was drawn in the first and in the second draw is: P(W3│(W1UW2)) =1/4Now, the probability of drawing 3 white balls one by one with out replacement(taking all 3 balls at a time gets the same analysis and result) is:P(W1UW2UW3) =P(W1)∙P(W2│W1)∙P(W3│(W1UW2)) =(3/6)∙(2/5)∙(1/4) =1/20=0.05 =5%.
Yes, it certainly can if there is only one possible outcome. For instance, the probability of drawing a red ball from a bag containing nothing but red balls is equal to one.
1 in 45
The probability is 3/7 * 2/6 * 1/5 = 1/35.
Baseball, a pocket billiards game, uses 21 numbered balls following the same color scheme as the first 15, and a cue ball.
40/50
25.1%The probability of having at least two same number balls when 7 balls are drawn withreplacement from a box containing 75 numbered balls from 1 to 75 is calculated asfollows:We calculate the probability of drawing 7 times a ball (with replacement) and havingall different numbers. Then we subtract this value from 1. This will be the probabilityof all the possibilities of combinations of 7 numbers where at least two are repeated.1.- For the 1st ball the options are 75/75.2.- For the 2nd ball the options for not repeating the 1st number are 74/75.3.- For the 3rd ball the options for not repeating the previous two numbers are 73/75.4.- For the 4th ball the options for not repeating the previous three numbers 72/75.5.- For the 5th ball ........................................................................... 71/75.6.- For the 6th ball ........................................................................... 70/75.7.- For the 7th ball ........................................................................... 69/75.The probability of drawing seven balls (with replacement) having all different numbersis: 75/75(74/75)(73/75)(72/75)(71/75)(70/75)(69/75) = 0.749419476...The probability of having at least two same number balls when 7 balls are drawn with replacement from the given box is:P = 1 - 0.749419476... = 0.250580524...≈ 25.1%
The probability is 1 if you draw three balls without replacement. If only one draw, it is 3/5.
The probability of a simple event is the number of ways to succeed over the total possibilities. So, there are three white balls, and ten balls total. So: P(white) = 3/10
Assuming that you're only drawing two balls out of the bag, your chance of getting two reds would be 3/5 * 2/4 = 3/10, or 30%