a man takes a step forward with probability 0.4 and backward with probability 0.6.find the probability that at the end of 5 step, he is one step away from the staring point
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One step away means a step forward or a step backward
=> either he took 6 forward steps and 5 backward steps or 6backward steps and 5 forward steps
=> reqd. probability
= 11C6 * (0.4)^6 * (0.5)^5 + 11C6 * (0.4)^5 * (0.6)^5
= 11C5 * (0.4)^5 * (0.6)^5 * (0.4 + 0.6) ... [11C6 = 11C5]
= (11x10x9x8x7)/(1x2x3x4x5) * (0.24)^5
= 0.368.Thanks and RegardsShaik AasifaskIITians faculty
=> either he took 6 forward steps and 5 backward steps or 6backward steps and 5 forward steps
=> reqd. probability
= 11C6 * (0.4)^6 * (0.5)^5 + 11C6 * (0.4)^5 * (0.6)^5
= 11C5 * (0.4)^5 * (0.6)^5 * (0.4 + 0.6) ... [11C6 = 11C5]
= (11x10x9x8x7)/(1x2x3x4x5) * (0.24)^5
= 0.368.Thanks and RegardsShaik AasifaskIITians faculty
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in 5 step s 1 step equal to 0.2 as probability so 06.-0.4=0.2
or 0.4-0.6= -0.2 so probability is 1
or 0.4-0.6= -0.2 so probability is 1
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