A jar contains 50 M&Ms.; 20 are red, 15 are green, 10 are yellow, and 5 are oran

Question

A jar contains 50 M&Ms.; 20 are red, 15 are green, 10 are yellow, and 5 are orange. You will draw M&Ms; at random from the jar, but this time without replacement. Find the following probabilities a) You draw four M&Ms; and they are all red. b) You draw three M&Ms; from the original 50 - first a green M&M;, then an orange M&M; and then another orange M&M; c) You draw out six M&Ms; and none of the M&Ms; is yellow d) You draw one M&Ms; and it is either red or green.

Explanation / Answer

N = Total number of M & Ms in Jar =50

R = number of red M & Ms. = 20

G = Number of Green M&Ms = 15

Y = Number of Yellow M&Ms = 10

O = Number of Orange M&Ms = 5

Since we draw M&Ms from the jar at random without replacement. That is we draw first M&Ms from jar at random and never replaced in the jar before the next drawing. The size of jar is reduced by one at every draw.

a) P( All four M &Ms are red) = P(First M&Ms are red) * P(Second M&Ms are red) * ( P(Third M&Ms are red) * P(fourth M&Ms are red)

= 20/50 * 19/49 * 18/48 *17/47

= 0.0210

b) P( First a green M&Ms, Second an orange M&Ms, Third is an orange M&Ms)

= P(First a green M&Ms) * P(second an orange M&Ms) * P( Third is an orange M&Ms.)

= 10/50 * 5/49 *4/48

= 0.0017

c) number of non-yellow M&Ms = 40

P( draw Six M&Ms and none of them yellow M&Ms) = 40/50 * 39/49 * 38/48 * 37/47 * 36/46 * 35/45

= 0.2415

d) P( Draw one M&Ms is either red or green ) = P( Draw one M&Ms is red) + P( Draw one M&Ms is green)

= 20/50 + 15/50

= 35/50

=0.70

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